Concise Mathematics Class 10 ICSE Solutions for Chapter 9 - Matrices

Exercise-9 A

1. State, whether the following statements are true or false. If false, give a reason.

(i) If A and B are two matrices of orders $\mathbf{\text{3 }\!\!\times\!\!\text{ 2}}$ and $\mathbf{2\times 3}$ respectively; then their sum A+B is possible.

Ans:  The two matrices are compatible for addition or subtraction; only when they have the same order. But the given matrices have different order. 

Therefore, the given statement is false.

(ii) The matrices and $\mathbf{{{B}_{2\times 3}}}$ are conformable for subtraction.

Ans:  The two matrices are compatible for addition or subtraction; only when they have the same order. 

The given matrices have the same order. 

Therefore, the given statement is true.

(iii) Transpose of a $\mathbf{2\times 1 matrix \;is \;a\; 2\times 1}$ matrix

Ans:  Transpose of a matrix of order $m\times n$ is $n\times m$ order. So, the transpose of a $2\times 1$ matrix is $1\times 2$.

Therefore, the given statement is false.

(iv) Transpose of a square matrix is a square matrix.

Ans: Transpose of any square matrix of order m×m is m×m order. 

Therefore, the given statement is true.

(v) A column matrix has many column and only one row.

Ans:  As the name suggest, column matrix has only one column. Therefore, the given statement is false.

2. Given: $\mathbf{\text{ }\!\![\!\!\text{ }\begin{matrix} \text{x} & \text{y+2} & \text{3} & \text{z-1}  \\ \end{matrix}\text{ }\!\!]\!\!\text{ }\,\text{=}\,\text{ }\!\![\!\!\text{ }\begin{matrix} \text{3} & \text{1} & \text{3} & \text{1}  \\ \end{matrix}\text{ }\!\!]\!\!\text{ }}$; find $\mathbf{x,~y~and~z}$.

Ans:

Given $[\begin{matrix} x & y+2 & 3 & z-1  \\ \end{matrix}]\,=\,[\begin{matrix} 3 & 1 & 3 & 1  \\ \end{matrix}]$ 

If two matrices are equal, their corresponding elements are also equal. Therefore,

x = 3 

$y+2=1\Rightarrow y=-1$ 

$z-1=2\Rightarrow z=3$ 

Therefore, the value of variables $x,~y~and~z$ are 3, -1 and 3 respectively.

3. Solve for a, b and c ; if:

(i) $[\begin{matrix} \mathbf{-4} & \mathbf{a+5} & \mathbf{3} &\mathbf{ 2} \\ \end{matrix}]\,\,=\,\,[\begin{matrix} \mathbf{ b+4} &\mathbf{ 2} & \mathbf{3} &\mathbf{ C-1} \\ \end{matrix}]$

Ans:

Given, $[\begin{matrix} -4 & a+5 & 3 & 2  \\ \end{matrix}]\,\,=\,\,[\begin{matrix}   b+4 & 2 & 3 & C-1  \\ \end{matrix}]$

If two matrices are equal, their corresponding elements are also equal. 

Therefore

$-4=b+4\Rightarrow b=-8$ 

$a+5=2\Rightarrow a=-3$ 

$c-1=2\Rightarrow c=3$ 

Therefore the value of variables $a,~b~and~c$ are -3, -8 and 3 respectively.

(ii)$[\begin{matrix} \mathbf{a} & \mathbf{a-b} & \mathbf{b+c} & \mathbf{0} \\ \end{matrix}]\,\,\text{=}\,\,[\begin{matrix} \mathbf{3} & \mathbf{-1} & \mathbf{2} & \mathbf{0} \\ \end{matrix}]$

Ans:

Given, $[\begin{matrix} a & a-b & b+c & 0  \\ \end{matrix}]\,\,=\,\,[\begin{matrix}  3 & -1 & 2 & 0  \\ \end{matrix}]$ 

If two matrices are equal, their corresponding elements are also equal. 

Therefore

a = 3

$a-b=-1\Rightarrow b=a+1$ 

b = 4 

$b+c=2\Rightarrow c=2-b$ 

c = -2

Therefore the value of variables $a,~b~and~c$ are 3, 4 and -2 respectively.

4. If $\mathbf{A}=\left[ \begin{matrix} \mathbf{8} & \mathbf{-3} \\ \end{matrix} \right]$ and $\mathbf{B}=\left[ \begin{matrix} \mathbf{4} & \mathbf{-5} \\ \end{matrix} \right]$ find

(i)A+B

Ans:

$\mathrm{A=}\left[ \begin{matrix} 8 & -3  \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 4 & -5  \\ \end{matrix} \right]$ 

$A+B=\left[ \begin{matrix} 8 & -3  \\ \end{matrix} \right]+\left[ \begin{matrix} 4 & -5  \\ \end{matrix} \right]$ 

$=\left[ \begin{matrix} 8+4 & -3-5  \\ \end{matrix} \right]$ 

$=\left[ \begin{matrix} 12 & -8  \\ \end{matrix} \right]$ 

(ii) B-A

Ans:

$\mathrm{A=}\left[ \begin{matrix} 8 & -3  \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix}4 & -5  \\ \end{matrix} \right]$

$\mathrm{B-A=}\left[ \mathrm{4 }\!\!~\!\!\text{ -5 }\!\!~\!\!\text{ } \right]\mathrm{-}\left[ \mathrm{8 }\!\!~\!\!\text{ -3 }\!\!~\!\!\text{ } \right]$ 

$\mathrm{=}\left[ \mathrm{4-8 }\!\!~\!\!\text{ }\,\,\,\,\mathrm{-5+3 }\!\!~\!\!\text{ } \right]$ 

$\mathrm{=}\left[ \mathrm{-4 }\!\!~\!\!\text{ }\,\,\,\mathrm{-2 }\!\!~\!\!\text{ } \right]$ 

5. If $\mathbf{A}=\begin{matrix} [\mathbf{2} & \mathbf{5} \\ \end{matrix}\left] ,~\mathbf{B}=\begin{matrix} [\mathbf{1} & \mathbf{4} \\ \end{matrix} \right]\text{ and}\; \mathbf{C= }\!\!~\!\!\text{ }\begin{matrix}[\mathbf{ 6} & \mathbf{-2} \\ \end{matrix}\text{ }\!\!]\!\!\text{ }$, find:

(i)B+C

Ans:

Given, $A=\begin{matrix} [2 & 5  \\ \end{matrix}\left] ,~B=\begin{matrix} [1 & 4  \\ \end{matrix} \right]$ and 

$C=\begin{matrix} [6 & -2  \\ \end{matrix}]$ 

$\mathrm{B+C=}\begin{matrix} \mathrm{ }\!\![\!\!\text{ 1} & \mathrm{4}  \\ \end{matrix}\mathrm{ }\!\!]\!\!\text{ +}$ $\begin{matrix} \mathrm{ }\!\![\!\!\text{ 6} & \mathrm{-2}  \\ \end{matrix}\mathrm{ }\!\!]\!\!\text{ }$ $\mathrm{=}\begin{matrix}  \mathrm{ }\!\![\!\!\text{ 1+6} & \mathrm{4-2}  \\ \end{matrix}\left] \mathrm{=}\begin{matrix} \mathrm{ }\!\![\!\!\text{ 7} & \mathrm{2}  \\ \end{matrix} \right]\mathrm{ }\!\!~\!\!\text{ }$

(ii) A-C 

Ans:

Given, $A=\begin{matrix} [2 & 5  \\ \end{matrix}\left] ,~B=\begin{matrix} [1 & 4  \\ \end{matrix} \right]$ and $C=\begin{matrix} [6 & -2  \\ \end{matrix}]$

A-C = $\left[ \begin{matrix} 2 & 5  \\ \end{matrix} \right]$- $\left[ \begin{matrix} 6 & -2  \\ \end{matrix} \right]$   

= $\left[ \begin{matrix} \mathrm{2-}\,\mathrm{6} & \mathrm{5}  \\ \end{matrix}\mathrm{+2} \right]\mathrm{=}\left[ \begin{matrix} \mathrm{-4} & \mathrm{7}  \\ \end{matrix} \right]$

(iii) A+B-C 

Ans:

Given, $A=\begin{matrix} [2 & 5  \\ \end{matrix}\left] ,~B=\begin{matrix} [1 & 4  \\ \end{matrix} \right]$ and $C=\begin{matrix} [6 & -2  \\ \end{matrix}]$

$A+B-C=\left[ \begin{matrix} \text{2} & \text{5}  \\ \end{matrix} \right]+\left[ \begin{matrix} \text{1} & \text{4}  \\ \end{matrix} \right]-\left[ \begin{matrix}    \text{6} & \text{-2}  \\ \end{matrix} \right]$  

$=\left[ \begin{matrix}    \text{2+1-6} & \text{5+4+2}  \\\end{matrix} \right]=\left[ \begin{matrix}   \text{-3} & \text{11}  \\\end{matrix} \right]$  

(iv) $\mathbf{A}-\mathbf{B}+\mathbf{C}$ 

Ans:

Given, $A=\left[ \begin{matrix} \text{2} & \text{5}  \\ \end{matrix} \right],\,\text{B=}\left[ \begin{matrix} \text{1} & \text{4}  \\ \end{matrix} \right]\,\,and\,C=\left[ \begin{matrix}  \text{6} & \text{-2}  \\\end{matrix} \right]$ 

$A-B+C=\left[ \begin{matrix}   \text{2} & \text{5}  \\\end{matrix} \right]-\left[ \begin{matrix}   \text{1} & \text{4}  \\\end{matrix} \right]+\left[ \begin{matrix}   \text{6} & \text{-2}  \\\end{matrix} \right]$ 

$=\left[ \begin{matrix}   \text{2-1+6} & \text{5-4-2}  \\\end{matrix} \right]=\left[ \begin{matrix}   \text{7} & \text{-1}  \\\end{matrix} \right]$  

6. Wherever possible write each of the following as a single matrix.

(i)$ \left[ \begin{matrix} \mathbf{ 1} &\mathbf{ 2} & \mathbf{3} & \mathbf{4} \\ \end{matrix} \right]+\left[ \begin{matrix} \mathbf{ -1} & \mathbf{-2} &\mathbf{ 1} &\mathbf{ -7} \\\end{matrix} \right]$

Ans:

$\begin{align}  & \left[ \begin{matrix}   \text{1} & \text{2} & \text{3} & \text{4}  \\\end{matrix} \right]\text{+}\left[ \begin{matrix}   \text{-1} & \text{-2} & \text{1} & \text{-7}  \\\end{matrix} \right]=\left[ \begin{matrix}   \text{1-1} & \text{2-2} & \text{3+1} & \text{4-7}  \\\end{matrix} \right] \\  & \text{=}\,\left[ \begin{matrix}   \text{0} & \text{0} & \text{4} & \text{-3}  \\\end{matrix} \right] \\ \end{align}$

(ii) $\left[ \begin{matrix} \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} &\mathbf{ 7} \\\end{matrix} \right]+\left[ \begin{matrix} \mathbf{0} & \mathbf{2} &\mathbf{ 3} & \mathbf{6} &\mathbf{ -1} & \mathbf{0} \\\end{matrix} \right]$

Ans:

$\begin{align}  & \left[ \begin{matrix}   \text{2} & \text{3} & \text{4} & \text{5} & \text{6} & \text{7}  \\\end{matrix} \right]\text{+}\left[ \begin{matrix}   \text{0} & \text{2} & \text{3} & \text{6} & \text{-1} & \text{0}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{2-0} & \text{3-2} & \text{4-3} & \text{5-6} & \text{6+1} & \text{7-0}  \\\end{matrix} \right] \\ & \text{=}\,\,\left[ \begin{matrix}   \text{2} & \text{1} & \text{1} & \text{-1} & \text{7} & \text{7}  \\\end{matrix} \right] \\ \end{align}$ 

(iii) $\left[ \begin{matrix} \mathbf{0} & \text{1} & \mathbf{2} & \mathbf{4} & \mathbf{6} & \mathbf{7} \\\end{matrix} \right]\text{+}\left[ \begin{matrix} \mathbf{3} & \mathbf{4} & \mathbf{6} & \mathbf{8} \\\end{matrix} \right]$

Ans:

$\left[ \begin{matrix}   \text{0} & \text{1} & \text{2} & \text{4} & \text{6} & \text{7}  \\\end{matrix} \right]\text{+}\left[ \begin{matrix}   \text{3} & \text{4} & \text{6} & \text{8}  \\\end{matrix} \right]$ not possible

Two matrices with different order cannot be unified as one matrix by the operation of addition. 

7. Find $\mathbf{x}~\mathbf{and}~\mathbf{y}$ from the following equations:

(i) $\left[ \begin{matrix} \mathbf{ 5} & \mathbf{ 2} & \mathbf{ -1} & \mathbf{ y-1} \\\end{matrix} \right]-\left[ \begin{matrix} \mathbf{1} & \mathbf{ x-1} & \mathbf{2} & \mathbf{-3} \\\end{matrix} \right]=\left[ \begin{matrix} \mathbf{4} & \mathbf{7} & \mathbf{-3} & \mathbf{2} \\\end{matrix} \right]$

Ans:

$\left[ \begin{matrix}   \text{5} & \text{2} & \text{-1} & \text{y-1}  \\\end{matrix} \right]\text{-}\left[ \begin{matrix}   \text{1} & \text{x-1} & \text{2} & \text{-3}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{4} & \text{7} & \text{-3} & \text{2}  \\\end{matrix} \right]$

$\begin{align}  & \left[ \begin{matrix}   \text{5-1} & \text{2-(x-1)} & \text{-1-2} & \text{y-1-(-3)}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{4} & \text{7} & \text{-3} & \text{2}  \\\end{matrix} \right] \\  & \left[ \begin{matrix}   \text{4} & \text{3-x} & \text{-3} & \text{y+2}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{4} & \text{7} & \text{-3} & \text{2}  \\\end{matrix} \right] \\ \end{align}$ 

Now by comparing the corresponding elements we have,

$\left[ \begin{matrix}   \text{-1} & \text{0} & \text{2} & \text{-4}  \\\end{matrix} \right]$

Therefore the value of $\text{x }\!\!~\!\!\text{ and }\!\!~\!\!\text{ y}$ is $-4\text{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }0$ respectively.

(ii) $\left[ \begin{matrix} \mathbf{-8} & \mathbf{x} \\\end{matrix} \right] + \left[ \begin{matrix} \mathbf{y} & \mathbf{-2} \\\end{matrix} \right]\,\,\text{=}\,\left[ \begin{matrix} \mathbf{-3} & \mathbf{2} \\\end{matrix} \right]$

Ans:

$\left[ \begin{matrix}   \text{-8} & \text{x}  \\\end{matrix} \right]\text{+}\left[ \begin{matrix}   \text{y} & \text{-2}  \\\end{matrix} \right]\,\,\text{=}\,\left[ \begin{matrix}   \text{-3} & \text{2}  \\\end{matrix} \right]$ 

$\left[ \begin{matrix}   \text{y-8} & \text{x-2}  \\\end{matrix} \right]\,\,\text{=}\,\left[ \begin{matrix}   \text{-3} & \text{2}  \\\end{matrix} \right]$ 

By comparing the corresponding elements we have

$y-8=-3\Rightarrow y=5$ 

$x-2=2\Rightarrow x=4$ 

Therefore the value of $x~and~y$ are $4~and~5$ respectively.

8. Given: M = $\left[ \begin{matrix} 5 & -3 & -2 & 4  \\\end{matrix} \right]$, find its transpose matrix ${{M}^{t}}$. If possible, find:

(i) $\mathbf{M}+{{\mathbf{M}}^{\mathbf{t}}}$

Ans: 

Given: M = $\left[ \begin{matrix}   \text{5} & \text{-3} & \text{-2} & \text{4}  \\\end{matrix} \right]$ 

Transpose of matrix M is ${{M}^{t}}$

Mt = $\left[ \begin{matrix}   \text{5} & \text{-2} & \text{-3} & \text{4}  \\\end{matrix} \right]$

M + Mt = $\left[ \begin{matrix}   \text{5} & \text{-3} & \text{-2} & \text{4}  \\\end{matrix} \right]$+ $\left[ \begin{matrix}   \text{5} & \text{-2} & \text{-3} & \text{4}  \\\end{matrix} \right]$

$\begin{align}  & =\left[ \begin{matrix}   \text{5+5} & \text{-3-2} & \text{-2-3} & \text{4+4}  \\\end{matrix} \right] \\  & \text{=}\left[ \begin{matrix}   \text{10} & \text{-5} & \text{-5} & \text{8}  \\\end{matrix} \right] \\ \end{align}$

(ii) ${{\mathbf{M}}^{\mathbf{t}}}-\mathbf{M}$ 

Ans: 

Given:  

Transpose of matrix M is ${{\text{M}}^{\text{t}}}$

Mt = $\left[ \begin{matrix}   5 & -3 & -2 & 4  \\\end{matrix} \right]$

Mt-M= $\left[ \begin{matrix}   \text{5} & \text{-2} & \text{-3} & \text{4}  \\\end{matrix} \right]$- $\left[ \begin{matrix}   \text{5} & \text{-3} & \text{-2} & \text{4}  \\\end{matrix} \right]$

$=\left[ \begin{matrix}   \text{5-5} & \text{-2+3} & \text{-3+2} & \text{4-4}  \\\end{matrix} \right]$ $\text{=}\left[ \begin{matrix}   \text{0} & \text{1} & \text{-1} & \text{0}  \\\end{matrix} \right]$

9. Write the additive inverse of matrices A, B and C: where A=$\left[ \begin{matrix} \mathbf{ 6} & \mathbf{-5} \\\end{matrix} \right]$; B=$\left[ \begin{matrix} \mathbf{ -2} & \mathbf{0} &\mathbf{ 4} & \mathbf{-1} \\\end{matrix} \right]$and C=$\left[ \begin{matrix} \mathbf{-7} & \mathbf{4} \\\end{matrix} \right]$

Ans: Given, A=$\left[ \begin{matrix}   \text{6} & \text{-5}  \\\end{matrix} \right]$; B=$\left[ \begin{matrix}   \text{-2} & \text{0} & \text{4} & \text{-1}  \\\end{matrix} \right]$ and C=$\left[ \begin{matrix}   \text{-7} & \text{4}  \\\end{matrix} \right]$  

The additive inverse of any matrix is the negative of each element of that matrix. Therefore, Additive inverse of matrix A is –A.

-A = $\left[ \begin{matrix}   \text{-6} & \text{5}  \\\end{matrix} \right]$ 

Additive inverse of matrix B is –B

-B = $\left[ \begin{matrix}   \text{2} & \text{0} & \text{-4} & \text{1}  \\\end{matrix} \right]$ 

Additive inverse of matrix C is –C

-C = $\left[ \begin{matrix}   \text{7} & \text{-4}  \\\end{matrix} \right]$ 

10. Given, A = $\left[ \begin{matrix} \mathbf{ 2} & \mathbf{ -3} \\\end{matrix} \right]$, $B=\left[ \begin{matrix} \mathbf{ 0} & \mathbf{ 2 } \\\end{matrix} \right]$ and $C=\left[ \begin{matrix} \mathbf{ -1} &\mathbf{ 4} \\\end{matrix} \right]$; find the matrix X in each of the following: 

(i) $\mathbf{X}+\mathbf{B}=\mathbf{C}-\mathbf{A}$

Ans: 

Given, $\text{A=}\left[ \begin{matrix}   \text{2} & \text{-3}  \\\end{matrix} \right]$, $\text{B=}\left[ \begin{matrix}   \text{0} & \text{2}  \\\end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix}   \text{-1} & \text{4}  \\\end{matrix} \right]$X + B = C – A X = C – A – B = $\left[ \begin{matrix}   \text{-1} & \text{4}  \\\end{matrix} \right]-\left[ \begin{matrix}   \text{2} & \text{-3}  \\\end{matrix} \right]-\left[ \begin{matrix}   \text{0} & \text{2}  \\\end{matrix} \right]$

$\begin{align}  & =\left[ \begin{matrix}   \text{-1-2-0} & \text{4+3-2}  \\\end{matrix} \right] \\  & \text{=}\left[ \begin{matrix}   \text{-3} & \text{5}  \\\end{matrix} \right] \\ \end{align}$

(ii) $\mathbf{A}-\mathbf{X}=\mathbf{B}+\mathbf{C}$

Ans:

Given, $\text{A=}\left[ \begin{matrix}   \text{2} & \text{-3}  \\\end{matrix} \right]$, $\text{B=}\left[ \begin{matrix}   \text{0} & \text{2}  \\\end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix}   \text{-1} & \text{4}  \\\end{matrix} \right]$A – X = B + CX = A – B – C = $\left[ \begin{matrix}   \text{2} & \text{-3}  \\\end{matrix} \right]-\left[ \begin{matrix}   \text{0} & \text{2}  \\\end{matrix} \right]-\left[ \begin{matrix}   \text{-1} & \text{4}  \\\end{matrix} \right]$ 

$=\left[ \begin{matrix}   \text{2-0+1} & \text{-3-2-4}  \\\end{matrix} \right]$ $\text{=}\left[ \begin{matrix}   \text{3} & \text{-9}  \\\end{matrix} \right]$ 

11. Given $A=\left[ \begin{matrix}   -1 & 0 & 2 & -4  \\\end{matrix} \right]$ and $B=\left[ \begin{matrix}   3 & -3 & -2 & 0  \\\end{matrix} \right]$; find the matrix X in each of the following : 

(i) $\mathbf{A}+\mathbf{X}=\mathbf{B}$

Ans:

$\text{A=}\left[ \begin{matrix}   \text{-1} & \text{0} & \text{2} & \text{-4}  \\\end{matrix} \right]$ and $\text{B=}\left[ \begin{matrix}   \text{3} & \text{-3} & \text{-2} & \text{0}  \\\end{matrix} \right]$

$\text{A}+\text{X}=\text{B}$ 

$\text{X=B-A=}\left[ \begin{matrix}   \text{3} & \text{-3} & \text{-2} & \text{0}  \\\end{matrix} \right]\text{-}\left[ \begin{matrix}   \text{-1} & \text{0} & \text{2} & \text{-4}  \\\end{matrix} \right]$

$\begin{align}  & \text{= }\!\![\!\!\text{ }\begin{matrix}   \text{3-(-1)} & \text{-3-0} & \text{-2-2} & \text{0-(-4)}  \\\end{matrix}\text{ }\!\!]\!\!\text{ } \\  & \text{=}\left[ \begin{matrix}   \text{4} & \text{-3} & \text{-4} & \text{4}  \\\end{matrix} \right] \\ \end{align}$ 

(ii) A-X=B

Ans:

$\text{A=}\left[ \begin{matrix}   \text{-1} & \text{0} & \text{2} & \text{-4}  \\\end{matrix} \right]\,\,\text{and}\,\text{B=}\left[ \begin{matrix}   \text{3} & \text{-3} & \text{-2} & \text{0}  \\\end{matrix} \right]$ $\text{A}-\text{X}=\text{B}$ 

$\begin{align}  & \text{X=A-B=}\left[ \begin{matrix}   \text{-1} & \text{0} & \text{2} & \text{-4}  \\\end{matrix} \right]\text{-}\left[ \begin{matrix}   \text{3} & \text{-3} & \text{-2} & \text{0}  \\\end{matrix} \right] \\  & \text{= }\!\![\!\!\text{ }\begin{matrix}   \text{-1-3} & \text{0+3} & \text{2+2} & \text{-4-0}  \\\end{matrix}\text{ }\!\!]\!\!\text{ } \\  & \text{=}\left[ \begin{matrix}   \text{-4} & \text{3} & \text{4} & \text{-4}  \\\end{matrix} \right] \\ \end{align}$

(iii) $\mathbf{X}-\mathbf{B}=\mathbf{A}$ 

Ans:

Given $\text{A=}\left[ \begin{matrix}   \text{-1} & \text{0} & \text{2} & \text{-4}  \\\end{matrix} \right]\,\,\text{and}\,\text{B=}\left[ \begin{matrix}   \text{3} & \text{-3} & \text{-2} & \text{0}  \\\end{matrix} \right]$

$\text{X}-\text{B}=\text{A}$  

$\text{X=A+B=}\left[ \begin{matrix}   \text{-1} & \text{0} & \text{2} & \text{-4}  \\\end{matrix} \right]\text{-}\left[ \begin{matrix}   \text{3} & \text{-3} & \text{-2} & \text{0}  \\\end{matrix} \right]$

$\text{= }\!\![\!\!\text{ }\begin{matrix}   \text{-1+3} & \text{0-3} & \text{2-2} & \text{-4+0}  \\\end{matrix}\text{ }\!\!]\!\!\text{ }$$\text{=}\left[ \begin{matrix}   \text{2} & \text{-3} & \text{0} & \text{-4}  \\\end{matrix} \right]$

Exercise-9 B

1. Evaluate:

(i) $3\left[ \begin{matrix} \mathbf{ 5} &\mathbf{ -2} \\\end{matrix} \right]$

Ans: 

$\text{3}\left[ \begin{matrix}   \text{5} & \text{-2}  \\\end{matrix} \right]=\left[ \begin{matrix}   \text{3 }\!\!\times\!\!\text{ 5} & \text{3 }\!\!\times\!\!\text{ (-2)}  \\\end{matrix} \right]$

$=\left[ \begin{matrix}   \text{15} & \text{-6}  \\\end{matrix} \right]$

(ii) $7\left[ \begin{matrix} \mathbf{ -1} & \mathbf{2} &\mathbf{ 0} &\mathbf{ 1} \\\end{matrix} \right]$

Ans: 

$\text{7}\left[ \begin{matrix}   \text{-1} & \text{2} & \text{0} & \text{1}  \\\end{matrix} \right]=\left[ \text{7 }\!\!\times\!\!\text{ (}\begin{matrix}   \text{-1)} & \text{7 }\!\!\times\!\!\text{ 2} & \text{7 }\!\!\times\!\!\text{ 0} & \text{7 }\!\!\times\!\!\text{ 1}  \\\end{matrix} \right]\,\,$   $\text{=}\left[ \begin{matrix}   \text{-7} & \text{14} & \text{0} & \text{7}  \\\end{matrix} \right]$

(iii) $2\left[ \begin{matrix} \mathbf{ -1 }& \mathbf{0} &\mathbf{ 2 }& \mathbf{-3} \\\end{matrix} \right]+\left[ \begin{matrix} \mathbf{ 3} & \mathbf{3} & \mathbf{5} &\mathbf{ 0} \\\end{matrix} \right]$

Ans:

$\text{2}\left[ \begin{matrix}   \text{-1} & \text{0} & \text{2} & \text{-3}  \\\end{matrix} \right]\text{+}\left[ \begin{matrix}   \text{3} & \text{3} & \text{5} & \text{0}  \\\end{matrix} \right]$

$\text{=}\left[ \text{2 }\!\!\times\!\!\text{ (}\begin{matrix}   \text{-1)} & \text{2 }\!\!\times\!\!\text{ 0} & \text{2 }\!\!\times\!\!\text{ 2} & \text{2 }\!\!\times\!\!\text{ (-3)}  \\\end{matrix} \right]\,\text{+}\left[ \begin{matrix}   \text{3} & \text{3} & \text{5} & \text{0}  \\\end{matrix} \right]\,$

$\text{=}\left[ \begin{matrix}   \text{-2+3} & \text{0+3} & \text{4+5} & \text{-6+0}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{1} & \text{3} & \text{9} & \text{-6}  \\\end{matrix} \right]$

 (iv) $6\left[ \begin{matrix} \mathbf{ 3} & \mathbf{ -2} \\\end{matrix} \right]-2\left[ \begin{matrix} \mathbf{ -8} & \mathbf{ 1 } \\\end{matrix} \right]$

Ans:

$\text{6}\left[ \begin{matrix}   \text{3} & \text{-2}  \\\end{matrix} \right]\text{-2}\left[ \begin{matrix}   \text{-8} & \text{1}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{18} & \text{-12}  \\\end{matrix} \right]\text{-}\left[ \begin{matrix}   \text{-16} & \text{2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{18+16} & \text{-12-2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{34} & \text{-14}  \\\end{matrix} \right]$

2. Find $x$ and $y$ if:

(i)$3\left[ \begin{matrix} \mathbf{4} & \mathbf{x} \\\end{matrix} \right]+2\left[ \begin{matrix} \mathbf{ y} & \mathbf{-3} \\\end{matrix} \right]=\left[ \begin{matrix} \mathbf{10} & \mathbf{0} \\\end{matrix} \right]$

Ans: 

$3\left[ \begin{matrix}   4 & x  \\\end{matrix} \right]+2\left[ \begin{matrix}   y & -3  \\\end{matrix} \right]=\left[ \begin{matrix}   10 & 0  \\\end{matrix} \right]$ $\left[ \begin{matrix}   12 & 3x  \\\end{matrix} \right]+\left[ 2\begin{matrix}   y & -6  \\\end{matrix} \right]=\left[ \begin{matrix}   10 & 0  \\\end{matrix} \right]$ 

$\left[ \begin{matrix}   12+2y & 3x-6  \\\end{matrix} \right]=\left[ \begin{matrix}   10 & 0  \\\end{matrix} \right]$ 

Now by comparing the corresponding elements we have

$12+2y=10\Rightarrow 2y=10-12$ 

$2y=-2\Rightarrow y=-1$ 

$3x-6=0\Rightarrow x=2$ 

Therefore x = 2 and y = -1 

(ii)$x\left[ \begin{matrix} \mathbf{ -1} &\mathbf{ 2} \\\end{matrix} \right]-4\left[ \begin{matrix} \mathbf{ -2} & \mathbf{ y} \\\end{matrix} \right]=\left[ \begin{matrix} \mathbf{ 7} & \mathbf{ -8} \\\end{matrix} \right]$

Ans: 

$x\left[ \begin{matrix}  -1 & 2  \\\end{matrix} \right]-4\left[ \begin{matrix}   -2 & y  \\\end{matrix} \right]=\left[ \begin{matrix}   7 & -8  \\\end{matrix} \right]$

$\left[ \begin{matrix}   -x & 2x  \\\end{matrix} \right]-\left[ \begin{matrix}   -8 & 4y  \\\end{matrix} \right]=\left[ \begin{matrix}   7 & -8  \\\end{matrix} \right]$ 

$\left[ \begin{matrix}   -x+8 & 2x-4y  \\\end{matrix} \right]=\left[ \begin{matrix}   7 & -8  \\\end{matrix} \right]$ 

Now by comparing we have

$-x+8=7\Rightarrow x=8-7$  

$\therefore x=1$ 

$2x-4y=-8\Rightarrow 2-4y=-8$ 

$4y=10\Rightarrow y=\dfrac{10}{4}$ 

$\therefore y=\dfrac{5}{2}$ 

Therefore the value of $x=1$ and $y=\dfrac{5}{2}$. 

3. Given $A=\left[ \begin{matrix}   2 & 1 & 3 & 0  \\\end{matrix} \right],\,B=\left[ \begin{matrix}   1 & 1 & 5 & 2  \\\end{matrix} \right]$ and $C=\left[ \begin{matrix}   -3 & -1 & 0 & 0  \\\end{matrix} \right]$; find:

(i) $2\mathbf{A}-3\mathbf{B}+\mathbf{C}$.  

Ans:

Given, $\text{A=}\left[ \begin{matrix}   \text{2} & \text{1} & \text{3} & \text{0}  \\\end{matrix} \right]\text{,}\,\text{B=}\left[ \begin{matrix}   \text{1} & \text{1} & \text{5} & \text{2}  \\\end{matrix} \right]\text{ and C=}\left[ \begin{matrix}   \text{-3} & \text{-1} & \text{0} & \text{0}  \\\end{matrix} \right]$

$2A-3B+C=2\left[ 2~1~3~0~ \right]-3\left[ 1~1~5~2~ \right]+\left[ -3~-1~0~0~ \right]$ 

$\text{2A-3B+C=2}\left[ \begin{matrix}   \text{2} & \text{1} & \text{3} & \text{0}  \\\end{matrix} \right]-\text{3}\left[ \begin{matrix}   \text{1} & \text{1} & \text{5} & \text{2}  \\\end{matrix} \right]\,+\left[ \begin{matrix}   \text{-3} & \text{-1} & \text{0} & \text{0}  \\\end{matrix} \right]$

$=\left[ \begin{matrix}   \text{4} & \text{2} & \text{6} & \text{0}  \\\end{matrix} \right]-\left[ \begin{matrix}   \text{3} & \text{3} & \text{15} & \text{6}  \\\end{matrix} \right]+\left[ \begin{matrix}   \text{-3} & \text{-1} & \text{0} & \text{0}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{-2} & \text{-2} & \text{-9} & \text{-6}  \\\end{matrix} \right]$

(ii) $\mathbf{A}+2\mathbf{C}-\mathbf{B}$ 

Ans:

Given

$\text{A=}\left[ \begin{matrix}   \text{2} & \text{1} & \text{3} & \text{0}  \\\end{matrix} \right]\text{,}\,\text{B=}\left[ \begin{matrix}   \text{1} & \text{1} & \text{5} & \text{2}  \\\end{matrix} \right]\text{ and C=}\left[ \begin{matrix}   \text{-3} & \text{-1} & \text{0} & \text{0}  \\\end{matrix} \right]$

$\text{A+2C-B=}\left[ \begin{matrix}   \text{2} & \text{1} & \text{3} & \text{0}  \\\end{matrix} \right]-\text{2}\left[ \begin{matrix}   \text{-3} & \text{-1} & \text{0} & \text{0}  \\\end{matrix} \right]-\left[ \begin{matrix}   \text{1} & \text{1} & \text{5} & \text{2}  \\\end{matrix} \right]\,$

$=\left[ \begin{matrix}   \text{2} & \text{1} & \text{3} & \text{0}  \\\end{matrix} \right]-\left[ \begin{matrix}   \text{-6} & \text{-2} & \text{0} & \text{0}  \\\end{matrix} \right]+\left[ \begin{matrix}   \text{1} & \text{1} & \text{5} & \text{2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{2-6-1} & \text{1-2-1} & \text{3+0-5} & \text{0+0-2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{-5} & \text{-2} & \text{-2} & \text{-2}  \\\end{matrix} \right]$

4. If $\left[ \begin{matrix}   4 & -2 & 4 & 0  \\\end{matrix} \right]+3A=\left[ \begin{matrix}   -2 & -2 & 1 & -3  \\\end{matrix} \right]$; find A.

Ans: 

Given

$\left[ \begin{matrix}   \text{4} & \text{-2} & \text{4} & \text{0}  \\\end{matrix} \right]\text{+3A=}\left[ \begin{matrix}   \text{-2} & \text{-2} & \text{1} & \text{-3}  \\\end{matrix} \right]$ $\text{3A=}\left[ \begin{matrix}   \text{-2} & \text{-2} & \text{1} & \text{-3}  \\\end{matrix} \right]\text{-}\left[ \begin{matrix}   \text{4} & \text{-2} & \text{4} & \text{0}  \\\end{matrix} \right]$$\text{A=}\dfrac{\text{1}}{\text{3}}\left[ \begin{matrix}   \text{-2+4} & \text{-2+2} & \text{1-4} & \text{-3-0}  \\\end{matrix} \right]$

$\text{A=}\left[ \begin{matrix}   \dfrac{\text{-6}}{\text{3}} & \text{0} & \dfrac{\text{-3}}{\text{3}} & \dfrac{\text{-3}}{\text{3}}  \\\end{matrix} \right]$

$\text{A=}\left[ \begin{matrix}   \text{-2} & \text{0} & \text{-1} & \text{-1}  \\\end{matrix} \right]$

5. Given $A=\left[ \begin{matrix}   1 & 4 & 2 & 3  \\\end{matrix} \right]$ and $B=\left[ \begin{matrix}   -4 & -1 & -3 & 2  \\\end{matrix} \right]$  

Find the matrix 2A+B 

Ans:

$\text{2A+B}\,\text{=}\,\text{2}\left[ \begin{matrix}   \text{1} & \text{4} & \text{2} & \text{3}  \\\end{matrix} \right]+\left[ \begin{matrix}   \text{-4} & \text{-1} & \text{-3} & \text{2}  \\\end{matrix} \right]$ 

$=\left[ \begin{matrix}   \text{2} & \text{8} & \text{4} & \text{6}  \\\end{matrix} \right]+\left[ \begin{matrix}   \text{-4} & \text{-1} & \text{-3} & \text{2}  \\\end{matrix} \right]$

$=\left[ \begin{matrix}   \text{-2} & \text{7} & \text{1} & \text{4}  \\\end{matrix} \right]$

6. Find a matrix C such that $C+B\,=\left[ \begin{matrix}   0 & 0 & 0 & 0  \\\end{matrix} \right]$ 

Ans:

$\text{C+B}\,\text{=}\left[ \begin{matrix}   \text{0} & \text{0} & \text{0} & \text{0}  \\\end{matrix} \right]$ 

$\text{C=-B=-}\left[ \begin{matrix}   \text{-4} & \text{-1} & \text{-3} & \text{-2}  \\\end{matrix} \right]$

$\text{C=}\left[ \begin{matrix}   \text{4} & \text{1} & \text{3} & \text{2}  \\\end{matrix} \right]$

7. If $2\left[ \begin{matrix}   3 & x & 0 & 1  \\\end{matrix} \right]+3\left[ \begin{matrix}   1 & 3 & y & 2  \\\end{matrix} \right]=\left[ \begin{matrix}   z & -7 & 15 & 8  \\\end{matrix} \right]$; find values of $x,~y$ and z. 

Ans: 

Given 

$\text{2}\left[ \begin{matrix}   \text{3} & \text{x} & \text{0} & \text{1}  \\\end{matrix} \right]\text{+3}\left[ \begin{matrix}   \text{1} & \text{3} & \text{y} & \text{2}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{z} & \text{-7} & \text{15} & \text{8}  \\\end{matrix} \right]$ 

$\left[ \text{6+}\begin{matrix}   \text{3} & \text{2x+9} & \text{0+3y} & \text{2+6}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{z} & \text{-7} & \text{15} & \text{8}  \\\end{matrix} \right]$

$\left[ \begin{matrix}   \text{9} & \text{2x+9} & \text{3y} & \text{8}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{z} & \text{-7} & \text{15} & \text{8}  \\\end{matrix} \right]$

By comparing we get

$9=z$ 

$2x+9=-7\Rightarrow 2x=-7-9$ 

x = -8 

$3y=15\Rightarrow y=5$ 

Thus, values of x = -8, y = 5 and z = 9. 

8. Given $A=\left[ \begin{matrix}   -3 & 6 & 0 & -9  \\\end{matrix} \right]$ and ${{\mathbf{A}}^{\mathbf{t}}}$ is its transpose matrix. Find, 

(i) $2\mathbf{A}+3{{\mathbf{A}}^{\mathbf{t}}}$ 

Ans:

$\text{A=}\left[ \begin{matrix}   \text{-3} & \text{6} & \text{0} & \text{-9}  \\\end{matrix} \right]\Rightarrow {{\text{A}}^{\text{t}}}\text{=}\left[ \begin{matrix}   \text{-3} & \text{0} & \text{6} & \text{-9}  \\\end{matrix} \right]$ 

$\text{2A+3}{{\text{A}}^{\text{t}}}\text{=2}\left[ \begin{matrix}   \text{-3} & \text{6} & \text{0} & \text{-9}  \\\end{matrix} \right]\text{+3}\left[ \begin{matrix}   \text{-3} & \text{0} & \text{6} & \text{-9}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{-6} & \text{12} & \text{0} & \text{-18}  \\\end{matrix} \right]\text{+}\left[ \begin{matrix}   \text{-9} & \text{0} & \text{18} & \text{-27}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{-15} & \text{12} & \text{18} & \text{-45}  \\\end{matrix} \right]$

(ii) $2{{A}^{t}}-3A$

Ans:

$\text{A=}\left[ \begin{matrix}   \text{-3} & \text{6} & \text{0} & \text{-9}  \\\end{matrix} \right]\Rightarrow {{\text{A}}^{\text{t}}}\text{=}\left[ \begin{matrix}   \text{-3} & \text{0} & \text{6} & \text{-9}  \\\end{matrix} \right]$ 

$\text{2A-3A=2}\left[ \begin{matrix}   \text{-3} & \text{6} & \text{0} & \text{-9}  \\\end{matrix} \right]\text{-3}\left[ \begin{matrix}   \text{-3} & \text{6} & \text{0} & \text{-9}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{-6} & \text{12} & \text{0} & \text{-18}  \\\end{matrix} \right]\text{+}\left[ \begin{matrix}   \text{-9} & \text{18} & \text{0} & \text{-27}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{-15} & \text{18} & \text{12} & \text{-45}  \\\end{matrix} \right]$

(iii) $\dfrac{1}{2}A-\dfrac{1}{3}{{A}^{t}}$

Ans:

$\text{A=}\left[ \begin{matrix}   \text{-3} & \text{6} & \text{0} & \text{-9}  \\\end{matrix} \right]\Rightarrow {{\text{A}}^{\text{t}}}\text{=}\left[ \begin{matrix}   \text{-3} & \text{0} & \text{6} & \text{-9}  \\\end{matrix} \right]$

$\dfrac{\text{1}}{\text{2}}\text{A-}\dfrac{\text{1}}{\text{3}}{{\text{A}}^{\text{t}}}\text{=}\dfrac{\text{1}}{\text{2}}\left[ \begin{matrix}   \text{-3} & \text{6} & \text{0} & \text{-9}  \\\end{matrix} \right]\text{-}\dfrac{\text{1}}{\text{3}}\left[ \begin{matrix}   \text{-3} & \text{0} & \text{6} & \text{-9}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \dfrac{\text{-3}}{\text{2}} & \text{3} & \text{0} & \dfrac{\text{-9}}{\text{2}}  \\\end{matrix} \right]\text{-}\left[ \begin{matrix}   \text{-1} & \text{0} & \text{2} & \text{-3}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \dfrac{\text{-3}}{\text{2}}\text{+1} & \text{3-0} & \text{0-2} & \dfrac{\text{-9}}{\text{2}}\text{+3}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \dfrac{\text{-1}}{\text{2}} & \text{3} & \text{-2} & \dfrac{\text{-3}}{\text{2}}  \\\end{matrix} \right]$

(iv) ${{A}^{t}}-\dfrac{1}{3}A$

Ans: 

$\text{A=}\left[ \begin{matrix}   \text{-3} & \text{6} & \text{0} & \text{-9}  \\\end{matrix} \right]\Rightarrow {{\text{A}}^{\text{t}}}\text{=}\left[ \begin{matrix}   \text{-3} & \text{0} & \text{6} & \text{-9}  \\\end{matrix} \right]$

${{\text{A}}^{\text{t}}}\text{-}\dfrac{\text{1}}{\text{3}}\text{A=}\left[ \begin{matrix}   \text{-3} & \text{0} & \text{6} & \text{-9}  \\\end{matrix} \right]\text{-}\dfrac{\text{1}}{\text{3}}\left[ \begin{matrix}   \text{-3} & \text{6} & \text{0} & \text{-9}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{-3} & \text{0} & \text{6} & \text{-9}  \\\end{matrix} \right]\text{-}\left[ \begin{matrix}   \text{-1} & \text{2} & \text{0} & \text{-3}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{-2} & \text{-2} & \text{6} & \text{-6}  \\\end{matrix} \right]$

9. Given $A=\left[ \begin{matrix}   1 & 1 & -2 & 0  \\\end{matrix} \right]$ and $B=\left[ \begin{matrix}   2 & -1 & 1 & 1  \\\end{matrix} \right]$. Solve for matrix X:

(i) X+2A=B

Ans:

Given, $\text{A=}\left[ \begin{matrix}   \text{1} & \text{1} & \text{-2} & \text{0}  \\\end{matrix} \right]\,\,and\,\text{B=}\left[ \begin{matrix}   \text{2} & \text{-1} & \text{1} & \text{1}  \\\end{matrix} \right]$ 

$\text{X+2A=B}$ 

$X=B-\text{2}A=\left[ \begin{matrix}   \text{2} & \text{-1} & \text{1} & \text{1}  \\\end{matrix} \right]-\text{2}\left[ \begin{matrix}   \text{1} & \text{1} & \text{-2} & \text{0}  \\\end{matrix} \right]\,$

$X=\left[ \begin{matrix}   \text{2-2} & \text{-1-2} & \text{1+4} & \text{1-0}  \\\end{matrix} \right]$

$X=\left[ \begin{matrix}   \text{0} & \text{-3} & \text{5} & \text{1}  \\\end{matrix} \right]$

(ii) 3X+B+2A=0 

Ans:

Given, $\text{A=}\left[ \begin{matrix}   \text{1} & \text{1} & \text{-2} & \text{0}  \\\end{matrix} \right]\,\,and\,\text{B=}\left[ \begin{matrix}   \text{2} & \text{-1} & \text{1} & \text{1}  \\\end{matrix} \right]$

$3\text{X}+\text{B}+2\text{A}=0$ 

$\text{3X=-B-2A=-}\left[ \begin{matrix}  \text{2} & \text{-1} & \text{1} & \text{1}  \\\end{matrix} \right]\text{-2}\left[ \begin{matrix}   \text{1} & \text{1} & \text{-2} & \text{0}  \\\end{matrix} \right]$

$\text{X=}\dfrac{\text{1}}{\text{3}}\left[ \text{-}\left[ \begin{matrix}   \text{2} & \text{-1} & \text{1} & \text{1}  \\\end{matrix} \right]\text{-}\left[ \begin{matrix}   \text{2} & \text{2} & \text{-4} & \text{0}  \\\end{matrix} \right] \right]$

$\text{X=}\dfrac{\text{1}}{\text{3}}\left[ \begin{matrix}   \text{-2-2} & \text{1-2} & \text{-2+4} & \text{-1-0}  \\\end{matrix} \right]$

$\text{X=}\left[ \begin{matrix} \text{-}\dfrac{\text{4}}{\text{3}} & \text{-}\dfrac{\text{1}}{\text{3}} & \text{1} & \text{-}\dfrac{\text{1}}{\text{3}}  \\\end{matrix} \right]$

(iii) $3\mathbf{A}-2\mathbf{X}=\mathbf{X}-2\mathbf{B}$ 

Ans:

Given, $\text{A=}\left[ \begin{matrix}  \text{1} & \text{1} & \text{-2} & \text{0}  \\\end{matrix} \right]\,\,and\,\text{B=}\left[ \begin{matrix}  \text{2} & \text{-1} & \text{1} & \text{1}  \\\end{matrix} \right]$

$\text{3A-2X=X-2B}$ 

$\text{3A+2B=X+2X}$ 

$\text{3X=3A+2B}$  

$\text{3X=3A=2B=3}\left[ \begin{matrix}   \text{1} & \text{1} & \text{-2} & \text{0}  \\\end{matrix} \right]\,+\text{2}\left[ \begin{matrix}   \text{2} & \text{-1} & \text{1} & \text{1}  \\\end{matrix} \right]$

$\text{X=}\dfrac{\text{1}}{\text{3}}\left[ \begin{matrix}   3+4 & 3-2 & -6+2 & 0+2  \\\end{matrix} \right]$ $\text{X=}\left[ \begin{matrix}   \dfrac{7}{3} & \dfrac{1}{3} & -\dfrac{4}{3} & \dfrac{2}{3}  \\\end{matrix} \right]$ 

10. If $M=\left[ \begin{matrix}   0 & 1  \\\end{matrix} \right]$ and $N=\left[ \begin{matrix}   1 & 0  \\\end{matrix} \right]$, show that: $3M+5N=\left[ \begin{matrix}   5 & 3  \\\end{matrix} \right]$ 

Ans: 

Given, $\text{M=}\left[ \begin{matrix}   \text{0} & \text{1}  \\\end{matrix} \right]$, $\text{N=}\left[ \begin{matrix}   \text{1} & \text{0}  \\\end{matrix} \right]$

We have to show that, $\text{3M+5N=}\left[ \begin{matrix}   \text{5} & \text{3}  \\\end{matrix} \right]$ $\text{LHS=3M+5N}$

$\text{=3}\left[ \begin{matrix}   0 & 1  \\\end{matrix} \right]+5\left[ \begin{matrix}   1 & 0  \\\end{matrix} \right]$

$=\left[ \begin{matrix}   0+5 & 3+0  \\\end{matrix} \right]$

$\text{LHS=}\left[ \begin{matrix}   5 & 3  \\\end{matrix} \right]\text{=RHS}$ 

Hence proved.

11. If I is the unit matrix of order $2\times 2$; find the matrix M, such that :

(i) $M-2I=3\left[ \begin{matrix}   -1 & 0 & 4 & 1  \\\end{matrix} \right]$ 

Ans:

Given $\text{I=}\left[ \begin{matrix}   \text{1} & \text{0} & \text{0} & \text{1}  \\\end{matrix} \right]$ 

$\text{M-2I=3}\left[ \begin{matrix}   \text{-1} & \text{0} & \text{4} & \text{1}  \\\end{matrix} \right]$ 

$\text{M=3}\left[ \begin{matrix}   \text{-1} & \text{0} & \text{4} & \text{1}  \\\end{matrix} \right]\text{+2}\left[ \begin{matrix}   \text{1} & \text{0} & \text{0} & \text{1}  \\\end{matrix} \right]$ 

$\text{=}\left[ \begin{matrix}   \text{-3+2} & \text{0+0} & \text{12+0} & \text{3+2}  \\\end{matrix} \right]$

$\text{M=}\left[ \begin{matrix}   \text{-1} & \text{0} & \text{12} & \text{5}  \\\end{matrix} \right]$

(ii) $5M+3I=4\left[ \begin{matrix}   2 & -5 & 0 & -3  \\\end{matrix} \right]$ 

Ans: 

Given $\text{I=}\left[ \begin{matrix}   \text{1} & \text{0} & \text{0} & \text{1}  \\\end{matrix} \right]$ 

$\text{5M+3I=4}\left[ \begin{matrix}   \text{2} & \text{-5} & \text{0} & \text{-3}  \\\end{matrix} \right]$ 

$\text{5M=4}\left[ \begin{matrix}   \text{2} & \text{-5} & \text{0} & \text{-3}  \\\end{matrix} \right]-\text{3}\left[ \begin{matrix}   \text{1} & \text{0} & \text{0} & \text{1}  \\\end{matrix} \right]$ 

$\text{M=}\dfrac{\text{1}}{\text{5}}\left[ \begin{matrix}   \text{8-3} & \text{-20-0} & \text{0-0} & \text{-12-3}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \dfrac{\text{5}}{\text{5}} & \text{-}\dfrac{\text{20}}{\text{5}} & \dfrac{\text{0}}{\text{5}} & \text{-}\dfrac{\text{15}}{\text{5}}  \\\end{matrix} \right]$  

$\text{M=}\left[ \begin{matrix}   \text{1} & \text{-4} & \text{0} & \text{-3}  \\\end{matrix} \right]$  

(iii) If $\left[ \begin{matrix}   1 & 4 & -2 & 3  \\\end{matrix} \right]+2M=3\left[ \begin{matrix}   3 & 2 & 0 & -3  \\\end{matrix} \right]$, find the matrix M. 

Ans: 

Given 

$\left[ \begin{matrix}   \text{1} & \text{4} & \text{-2} & \text{3}  \\\end{matrix} \right]\text{+2M=3}\left[ \begin{matrix}   \text{3} & \text{2} & \text{0} & \text{-3}  \\\end{matrix} \right]$ 

$\text{2M=3}\left[ \begin{matrix}   \text{3} & \text{2} & \text{0} & \text{-3}  \\\end{matrix} \right]-\left[ \begin{matrix}   \text{1} & \text{4} & \text{-2} & \text{3}  \\\end{matrix} \right]$

$\text{M=}\dfrac{\text{1}}{\text{2}}\left[ \begin{matrix}   \text{9-1} & \text{6-4} & \text{0+2} & \text{-9-3}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix}   \dfrac{\text{8}}{\text{2}} & \dfrac{\text{2}}{\text{2}} & \dfrac{\text{2}}{\text{2}} & \text{-}\dfrac{\text{12}}{\text{2}}  \\\end{matrix} \right]$

$\text{M=}\left[ \begin{matrix}   \text{4} & \text{1} & \text{1} & \text{-6}  \\\end{matrix} \right]$

Exercise-9 C

1. Evaluate: if possible:

(i) $\left[ \begin{matrix}  \mathbf{ 3} & \mathbf{2}  \\\end{matrix} \right]\left[ \begin{matrix} \mathbf{  2}  &\mathbf{ 0}  \\\end{matrix} \right]$ 

Ans:

$\left[ \begin{matrix}   \text{3} & \text{2}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{2} & \text{0}  \\\end{matrix} \right]$$=\left[ 3\times 2+2\times 0 \right]$ 

$=\left[ 6 \right]$ 

(ii) $\left[ \begin{matrix} \mathbf{  1} &\mathbf{ -2}  \\\end{matrix} \right]\left[ \begin{matrix}   \mathbf{-2} &\mathbf{ 3} &\mathbf{ -1} & \mathbf{4}  \\\end{matrix} \right]$ 

Ans:

$\left[ \begin{matrix}   \text{1} & \text{-2}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{-2} & \text{3} & \text{-1} & \text{4}  \\\end{matrix} \right]=\left[ \text{1 }\!\!\times\!\!\text{ (-2)+(-2) }\!\!\times\!\!\text{ (-1) }\!\!\times\!\!\text{ 3+(-2) }\!\!\times\!\!\text{ 4} \right]$ 

$\text{=}\left[ \begin{matrix}   \text{0} & \text{-5}  \\\end{matrix} \right]$

(iii) $\left[ \begin{matrix}   \mathbf{6} & \mathbf{4} & \mathbf{3} & \mathbf{-1}  \\\end{matrix} \right]\left[ \begin{matrix}  \mathbf{ -1} &\mathbf{ 3}  \\\end{matrix} \right]$

Ans:

$\left[ \begin{matrix}   \text{6} & \text{4} & \text{3} & \text{-1}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{-1} & \text{3}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix} \text{6 }\!\!\times\!\!\text{ (-1)+4 }\!\!\times\!\!\text{ 3} & \text{3 }\!\!\times\!\!\text{ (-1)+(-1) }\!\!\times\!\!\text{ 3}  \\\end{matrix} \right]$  

$=\left[ \begin{matrix}   \text{6} & \text{-6}  \\\end{matrix} \right]$

(iv) $\left[ \begin{matrix}   \mathbf{6} &\mathbf{ 4 }&\mathbf{ 3} &\mathbf{ -1}  \\\end{matrix} \right]\left[ \begin{matrix}  \mathbf{ -1 }&\mathbf{ 3}  \\\end{matrix} \right]$

Ans:

$\left[ \begin{matrix}   \text{6} & \text{4} & \text{3} & \text{-1}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{-1} & \text{3}  \\\end{matrix} \right]$ 

Order of the first matrix is $2\times 2$ and the order of the second matrix is $1\times 2$.

The number of columns in the first matrix is not equal to the number of rows in the second matrix. Thus, the product is not possible.

2. If $A=\left[ \begin{matrix}  \mathbf{ 0} & \mathbf{2} &\mathbf{ 5 }&\mathbf{ -2}  \\\end{matrix} \right],\,B=\left[ \begin{matrix}  \mathbf{ 1} &\mathbf{ -1} &\mathbf{ 3} &\mathbf{ 2}  \\\end{matrix} \right]$ and I is a unit matrix of order $2\times 2$, find: 

(i) AB 

Ans: 

$\text{AB=}\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{1} & \text{-1} & \text{3} & \text{2}  \\\end{matrix} \right]$ 

$\text{AB=}\left[ \begin{matrix}   \text{0 }\!\!\times\!\!\text{ 1+2 }\!\!\times\!\!\text{ 3} & \text{0 }\!\!\times\!\!\text{ (-1)+2 }\!\!\times\!\!\text{ 2} & \text{5 }\!\!\times\!\!\text{ 1+(-2) }\!\!\times\!\!\text{ 3} & \text{5 }\!\!\times\!\!\text{ (-1)+(-2) }\!\!\times\!\!\text{ 2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{6} & \text{4} & \text{-1} & \text{-9}  \\\end{matrix} \right]$

 (ii) BA 

Ans: 

$\text{BA=}\left[ \begin{matrix}   \text{1} & \text{-1} & \text{3} & \text{2}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{1 }\!\!\times\!\!\text{ 0+(-1) }\!\!\times\!\!\text{ 5} & \text{1 }\!\!\times\!\!\text{ 2+(-1) }\!\!\times\!\!\text{ (-2)} & \text{3 }\!\!\times\!\!\text{ 0+2 }\!\!\times\!\!\text{ 5} & \text{3 }\!\!\times\!\!\text{ 2+2 }\!\!\times\!\!\text{ (-2)}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{-5} & \text{4} & \text{10} & \text{2}  \\\end{matrix} \right]$

(iii) AI 

Ans:

$\text{A=}\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right],\,I=\left[ \begin{matrix}   \text{1} & \text{0} & \text{0} & \text{1}  \\\end{matrix} \right]$

$\text{AI=}\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{1} & \text{0} & \text{0} & \text{1}  \\\end{matrix} \right]$

$\text{AI=}\left[ \begin{matrix}   \text{0 }\!\!\times\!\!\text{ 1+2 }\!\!\times\!\!\text{ 0} & \text{0 }\!\!\times\!\!\text{ 0+2 }\!\!\times\!\!\text{ 1} & \text{5 }\!\!\times\!\!\text{ 1+(-2) }\!\!\times\!\!\text{ 0} & \text{5 }\!\!\times\!\!\text{ 0+(-2) }\!\!\times\!\!\text{ 1}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right]$ 

(iv) IB 

Ans:

$\text{B=}\left[ \begin{matrix}   \text{1} & \text{-1} & \text{3} & \text{2}  \\\end{matrix} \right]\text{,}\,\text{I=}\left[ \begin{matrix}   \text{1} & \text{0} & \text{0} & \text{1}  \\\end{matrix} \right]$

$\text{IB=}\left[ \begin{matrix}   \text{1} & \text{0} & \text{0} & \text{1}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{1} & \text{-1} & \text{3} & \text{2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{1 }\!\!\times\!\!\text{ 1+0 }\!\!\times\!\!\text{ 3} & \text{1 }\!\!\times\!\!\text{ (-1)+0 }\!\!\times\!\!\text{ 2} & \text{0 }\!\!\times\!\!\text{ 1+1 }\!\!\times\!\!\text{ 3} & \text{0 }\!\!\times\!\!\text{ (-1)+1 }\!\!\times\!\!\text{ 2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{1} & \text{-1} & \text{3} & \text{2}  \\\end{matrix} \right]$

(v) ${{\mathbf{A}}^{2}}$

Ans:

$A=\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right]$${{A}^{2}}=\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{0 }\!\!\times\!\!\text{ 0+2 }\!\!\times\!\!\text{ 5} & \text{0 }\!\!\times\!\!\text{ 2+2 }\!\!\times\!\!\text{ (-2)} & \text{5 }\!\!\times\!\!\text{ 0+(-2) }\!\!\times\!\!\text{ 5} & \text{5 }\!\!\times\!\!\text{ 2+(-2) }\!\!\times\!\!\text{ (-2)}  \\\end{matrix} \right]$$\text{=}\left[ \begin{matrix}   \text{10} & \text{-4} & \text{-10} & \text{14}  \\\end{matrix} \right]$

(vi) ${{\mathbf{B}}^{2}}\mathbf{A}$ 

Ans:

$\text{B=}\left[ \begin{matrix}   \text{1} & \text{-1} & \text{3} & \text{2}  \\\end{matrix} \right],\,A=\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right]$ ${{B}^{2}}A=\left( \left[ \begin{matrix}   \text{1} & \text{-1} & \text{3} & \text{2}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{1} & \text{-1} & \text{3} & \text{2}  \\\end{matrix} \right] \right)\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{1 }\!\!\times\!\!\text{ 1+(-1) }\!\!\times\!\!\text{ 3} & \text{1 }\!\!\times\!\!\text{ (-1)+(-1) }\!\!\times\!\!\text{ 2} & \text{3 }\!\!\times\!\!\text{ 1+2 }\!\!\times\!\!\text{ 3} & \text{3 }\!\!\times\!\!\text{ (-1)+2 }\!\!\times\!\!\text{ 2}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{-2} & \text{-3} & \text{9} & \text{1}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{0} & \text{2} & \text{5} & \text{-2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}   \text{-2 }\!\!\times\!\!\text{ 0+(-3) }\!\!\times\!\!\text{ 5} & \text{-2 }\!\!\times\!\!\text{ 2+(-3) }\!\!\times\!\!\text{ (2)} & \text{9 }\!\!\times\!\!\text{ 0+1 }\!\!\times\!\!\text{ 5} & \text{9 }\!\!\times\!\!\text{ 2+1 }\!\!\times\!\!\text{ (-2)}  \\\end{matrix} \right]$

$=\left[ \begin{matrix}   \text{-15} & \text{2} & \text{5} & \text{16}  \\\end{matrix} \right]$

3. If $A=\left[ \begin{matrix}  \mathbf{ 3} & \mathbf{x }&\mathbf{ 0} &\mathbf{ 1}  \\\end{matrix} \right]$ and $B=\left[ \begin{matrix}   \mathbf{9} &\mathbf{ 16} & \mathbf{0} &\mathbf{ -y}  \\\end{matrix} \right]$, find x and y when ${{A}^{2}}=B$. 

Ans: 

Given $\text{A=}\left[ \begin{matrix}   \text{3} & \text{x} & \text{0} & \text{1}  \\\end{matrix} \right]$ and $B=\left[ \begin{matrix}   \text{9} & \text{16} & \text{0} & \text{-y}  \\\end{matrix} \right]$ 

${{\text{A}}^{2}}=\text{B}$ 

$\left[ \begin{matrix}   \text{3} & \text{x} & \text{0} & \text{1}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{3} & \text{x} & \text{0} & \text{1}  \\\end{matrix} \right]=\left[ \begin{matrix}   \text{9} & \text{16} & \text{0} & -y  \\\end{matrix} \right]$

$\left[ \begin{matrix}   \text{9+0} & \text{3x+x} & \text{0+0} & \text{0+1}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{9} & \text{16} & \text{0} & \text{-y}  \\\end{matrix} \right]$ 

By comparing we have

$4x=16\Rightarrow x=4$ 

$-y=1\Rightarrow y=-1$ 

Therefore the value of $x=4$ and $\text{y}=-1$. 

4. Find x and y if:

(i) $\left[ \begin{matrix} \mathbf{  4} &\mathbf{ 3x} & \mathbf{x }& \mathbf{-2}  \\\end{matrix} \right]\left[ \begin{matrix}  \mathbf{ 5} &\mathbf{ 1}  \\\end{matrix} \right]\,=\left[ \begin{matrix}   \mathbf{y} &\mathbf{ 8}  \\\end{matrix} \right]\,\,$ 

Ans:

$\left[ \begin{matrix}   \text{4} & \text{3x} & \text{x} & \text{-2}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{5} & \text{1}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{y} & \text{8}  \\\end{matrix} \right]$ 

$\left[ \begin{matrix}   \text{20+3x} & \text{5x-2}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{y} & \text{8}  \\\end{matrix} \right]$ 

By comparing we have

$20+3x=y$  and $5x-2=8\Rightarrow 5x=10$ 

$\therefore x=2$ 

$y=20+6=26$ 

Thus value of $x=2$ and $y=26$ 

(ii) $\left[ \begin{matrix}   \mathbf{x} & \mathbf{0 }& \mathbf{-3} & \mathbf{1}  \\\end{matrix} \right]\left[ \begin{matrix}  \mathbf{ 1} &\mathbf{ 1} & \mathbf{0} & \mathbf{y}  \\\end{matrix} \right]=\left[ \begin{matrix}  \mathbf{ 2} &\mathbf{ 2} &\mathbf{ -3} & \mathbf{-2}  \\\end{matrix} \right]$ 

Ans:

$\left[ \begin{matrix}   \text{x} & \text{0} & \text{-3} & \text{1}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{1} & \text{1} & \text{0} & \text{y}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{2} & \text{2} & \text{-3} & \text{-2}  \\\end{matrix} \right]$

$\left[ \begin{matrix}   \text{x} & \text{x} & \text{-3} & \text{y-3}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{2} & \text{2} & \text{-3} & \text{-2}  \\\end{matrix} \right]$ 

By comparing we have,

x=2 and y-3=-2 

y=1

Thus, value of x=2 and y=1. 

5. If $A=\left[ \begin{matrix}  \mathbf{ 1} & \mathbf{3} & \mathbf{2} & \mathbf{4}  \\\end{matrix} \right],\,B=\left[ \begin{matrix}  \mathbf{ 1} &\mathbf{ 2} & \mathbf{4} & \mathbf{3 } \\\end{matrix} \right]$ and $C=\left[ \begin{matrix} \mathbf{  4} &\mathbf{ 3} &\mathbf{ 1 }& \mathbf{2}  \\\end{matrix} \right]$, find:

(i) $\left( \mathbf{AB} \right)\mathbf{C}$

Ans:

Given, $\text{A=}\left[ \begin{matrix}   \text{1} & \text{3} & \text{2} & \text{4}  \\\end{matrix} \right]\text{,}\,\text{B=}\left[ \begin{matrix}   \text{1} & \text{2} & \text{4} & \text{3}  \\\end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix}  \text{4} & \text{3} & \text{1} & \text{2}  \\\end{matrix} \right]$

$(AB)C=\left( \left[ \begin{matrix}   \text{1} & \text{3} & \text{2} & \text{4}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{1} & \text{2} & \text{4} & \text{3}  \\\end{matrix} \right] \right)\left[ \begin{matrix}   \text{4} & \text{3} & \text{1} & \text{2}  \\\end{matrix} \right]$

$\text{=}\left( \left[ \begin{matrix}   \text{13} & \text{11} & \text{18} & \text{16}  \\\end{matrix} \right] \right)\left[ \begin{matrix}   \text{4} & \text{3} & \text{1} & \text{2}  \\\end{matrix} \right]$ 

$=\left[ \begin{matrix}   \text{63} & \text{61} & \text{88} & \text{86}  \\\end{matrix} \right]$ 

(ii) A(BC)

Ans:

Given, $\text{A=}\left[ \begin{matrix}   \text{1} & \text{3} & \text{2} & \text{4}  \\\end{matrix} \right]\text{,}\,\text{B=}\left[ \begin{matrix}   \text{1} & \text{2} & \text{4} & \text{3}  \\\end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix}   \text{4} & \text{3} & \text{1} & \text{2}  \\\end{matrix} \right]$

$A(BC)=\left[ \begin{matrix}   \text{1} & \text{3} & \text{2} & \text{4}  \\\end{matrix} \right]\left( \left[ \begin{matrix}   \text{1} & \text{2} & \text{4} & \text{3}  \\\end{matrix} \right]\left[ \begin{matrix}   \text{4} & \text{3} & \text{1} & \text{2}  \\\end{matrix} \right] \right)$ 

$\text{=}\left[ \begin{matrix}   \text{1} & \text{3} & \text{2} & \text{4}  \\\end{matrix} \right]\left( \left[ \begin{matrix}   \text{6} & \text{7} & \text{19} & \text{18}  \\\end{matrix} \right] \right)$

$\text{=}\left[ \begin{matrix}   \text{63} & \text{61} & \text{88} & \text{86}  \\\end{matrix} \right]$ 

yes, (AB)C=A(BC). 

6. Given $A=\left[ \begin{matrix} \mathbf{0} & \mathbf{4 }& \mathbf{6} &\mathbf{ 3} &\mathbf{ 0} &\mathbf{ -1}  \\\end{matrix} \right]$ and $B=\left[ \begin{matrix} \mathbf{0} &\mathbf{ 1} & \mathbf{-1} & \mathbf{2} & \mathbf{-5} & \mathbf{-6}  \\ \end{matrix} \right]$, is the following possible:

(i) $\mathbf{AB}$

Ans:

Given, $\left[ \begin{matrix} \text{0} & \text{4} & \text{6} & \text{3} & \text{0} & \text{-1}  \\ \end{matrix} \right]$ and $\left[ \begin{matrix} \text{0} & \text{1} & \text{-1} & \text{2} & \text{-5} & \text{-6}  \\\end{matrix} \right]$  

$\text{AB=}\left[ \begin{matrix} \text{0} & \text{4} & \text{6} & \text{3} & \text{0} & \text{-1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{0} & \text{1} & \text{-1} & \text{2} & \text{-5} & \text{-6}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix} \text{0-4-30} & \text{0+8-36} & \text{0+0+5} & \text{3+0+6}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix} \text{-34} & \text{28} & \text{5} & \text{9}  \\\end{matrix} \right]$

(ii) BA

Ans:

Given, $\left[ \begin{matrix} \text{0} & \text{4} & \text{6} & \text{3} & \text{0} & \text{-1}  \\ \end{matrix} \right]$ and $\left[ \begin{matrix} \text{0} & \text{1} & \text{-1} & \text{2} & \text{-5} & \text{-6}  \\ \end{matrix} \right]$ 

$\text{BA=}\left[ \begin{matrix} \text{0} & \text{1} & \text{-1} & \text{2} & \text{-5} & \text{-6}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{0} & \text{4} & \text{6} & \text{3} & \text{0} & \text{-1}  \\ \end{matrix} \right]$

$\text{=}\left[ \begin{matrix} \text{0+3} & \text{0+0} & \text{0-1} & \text{0+6} & \text{-4+0} & \text{-6-2} & \text{0-18} & \text{-20+0} & \text{-30+6}  \\ \end{matrix} \right]$

$\text{=}\left[ \begin{matrix} \text{3} & \text{0} & \text{-1} & \text{6} & \text{-4} & \text{-8} & \text{-18} & \text{-20} & \text{-24}  \\\end{matrix} \right]$

(iii) A2

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{0} & \text{4} & \text{6} & \text{3} & \text{0} & \text{-1}  \\ \end{matrix} \right]$

A2 = $\left[ \begin{matrix} \text{0} & \text{4} & \text{6} & \text{3} & \text{0} & \text{-1}  \\ \end{matrix} \right]$$\left[ \begin{matrix} \text{0} & \text{4} & \text{6} & \text{3} & \text{0} & \text{-1}  \\ \end{matrix} \right]$

The number of columns of matrix A is not equal to its number of rows. Therefore, product A2 is not possible. 

7. Let $A=\left[ \begin{matrix} \mathbf{2} & \mathbf{1} & \mathbf{0} & \mathbf{-2}  \\\end{matrix} \right],\,B=\left[ \begin{matrix} \mathbf{4} & \mathbf{1} &\mathbf{ -3} & \mathbf{-2}  \\ \end{matrix} \right]$ and $C=\left[ \begin{matrix} \mathbf{-3} & \mathbf{2} & \mathbf{-1} & \mathbf{4}  \\ \end{matrix} \right]$. Find ${{\text{A}}^{\text{2}}}\text{+AC-5B}$.

Solutions:

Given $\text{A=}\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{-2}  \\\end{matrix} \right]\text{,}\,\text{B=}\left[ \begin{matrix} \text{4} & \text{1} & \text{-3} & \text{-2}  \\\end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix} \text{-3} & \text{2} & \text{-1} & \text{4}  \\ \end{matrix} \right]$

${{\text{A}}^{\text{2}}}\text{+AC-5B=}\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{-2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{-2}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{-2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{-3} & \text{2} & \text{-1} & \text{4}  \\ \end{matrix} \right]\text{-5}\left[ \begin{matrix} \text{4} & \text{1} & \text{-3} & \text{-2}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix} \text{4} & \text{0} & \text{0} & \text{4}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{-7} & \text{8} & \text{2} & \text{-8}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{20} & \text{5} & \text{-15} & \text{-10}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix} \text{4-7-20} & \text{0+8-5} & \text{0+2+15} & \text{4-8+10}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix}\text{-23} & \text{3} & \text{17} & \text{6}  \\\end{matrix} \right]$

8. If $M=\left[ \begin{matrix} \mathbf{1} & \mathbf{2} &\mathbf{ 2} &\mathbf{ 1}  \\\end{matrix} \right]$ and I is a unit matrix of the same order as that of M; show that: 

(i) ${{M}^{2}}=2M+3I$ 

Ans:

Given, $\text{M=}\left[ \begin{matrix} \text{1} & \text{2} & \text{2} & \text{1}  \\\end{matrix} \right]$

${{\text{M}}^{2}}=2\text{M}+3\text{I}$ 

$\text{LHS}={{\text{M}}^{2}}$ 

$\text{=}\left[ \begin{matrix} \text{1} & \text{2} & \text{2} & \text{1}  \\\end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{2} & \text{2} & \text{1}  \\\end{matrix} \right]$

$\text{=}\left[ \begin{matrix} \text{5} & \text{4} & \text{4} & \text{5}  \\\end{matrix} \right]$

$\text{RHS}=2\text{M}+3\text{I}$ 

$\begin{align} & \text{=2}\left[ \begin{matrix} \text{1} & \text{2} & \text{2} & \text{1}  \\ \end{matrix} \right]\text{+3}\left[ \begin{matrix}\text{1} & \text{0} & \text{0} & \text{1}  \\\end{matrix} \right] \\ & \text{=}\left[ \begin{matrix} \text{2+3} & \text{4+0} & \text{4+0} & \text{2+3}  \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \text{5} & \text{4} & \text{4} & \text{5}  \\\end{matrix} \right] \\  \end{align}$ = LHS

Hence proved. 

9. If $A=\left[ \begin{matrix} \mathbf{a} & \mathbf{0} & \mathbf{0} & \mathbf{2}  \\ \end{matrix} \right],\,B=\left[ \begin{matrix} \mathbf{0} & \mathbf{-b} & \mathbf{1} &\mathbf{ 0}  \\ \end{matrix} \right],\,M=\left[ \begin{matrix}\mathbf{ 1} & \mathbf{-1} & \mathbf{1 }& \mathbf{1}  \\ \end{matrix} \right]$ and $\mathbf{BA}={{\mathbf{M}}^{2}}$, find the values of a and b.

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{a} & \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]\text{,}\,\text{B=}\left[ \begin{matrix} \text{0} & \text{-b} & \text{1} & \text{0}  \\ \end{matrix} \right]\text{,}\,\text{M=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{1} & \text{1}  \\ \end{matrix} \right]$

Also $\text{BA}={{\text{M}}^{2}}$

$\begin{align} & \left[ \begin{matrix} \text{0} & \text{-b} & \text{1} & \text{0}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{a} & \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]\,\text{=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{1} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{-1} & \text{1} & \text{1}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{0} & \text{-2b} & \text{a} & \text{0}  \\ \end{matrix} \right]=\left[ \begin{matrix} \text{0} & \text{-2} & \text{2} & \text{0}  \\ \end{matrix} \right] \\ \end{align}$

By comparing we have

$-2b=-2\therefore b=1$ 

a = 2 

Hence the value of a = 2 and b = 1. 

10. Given $A=\left[ \begin{matrix}\mathbf{ 4} &\mathbf{ 1} & \mathbf{2 }& \mathbf{3 } \\\end{matrix} \right]$ and $B=\left[ \begin{matrix} \mathbf{1} &\mathbf{ 0} &\mathbf{ -2} & \mathbf{1}  \\ \end{matrix} \right]$, find:

(i) $\mathbf{A}-\mathbf{B}$

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{4} & \text{1} & \text{2} & \text{3}  \\ \end{matrix} \right]\,\,and\,\text{B=}\left[ \begin{matrix} \text{1} & \text{0} & \text{-2} & \text{1}  \\ \end{matrix} \right]$ 

$\begin{align} & \text{A-B=}\left[ \begin{matrix} \text{4} & \text{1} & \text{2} & \text{3}  \\ \end{matrix} \right]\,-\left[ \begin{matrix} \text{1} & \text{0} & \text{-2} & \text{1}  \\ \end{matrix} \right]\, \\  & =\left[ \begin{matrix} \text{3} & \text{1} & \text{4} & \text{2}  \\\end{matrix} \right] \\ \end{align}$

(ii) ${{\mathbf{A}}^{2}}$

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{4} & \text{1} & \text{2} & \text{3}  \\\end{matrix} \right]$

$\begin{align} & {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{4} & \text{1} & \text{2} & \text{3}  \\\end{matrix} \right]\left[ \begin{matrix} \text{4} & \text{1} & \text{2} & \text{3}  \\\end{matrix} \right]\,\, \\  & \text{=}\left[ \begin{matrix} \text{18} & \text{7} & \text{14} & \text{11}  \\\end{matrix} \right] \\ \end{align}$

(iii) $\mathbf{AB}$ 

Ans: 

Given $\text{A=}\left[ \begin{matrix} \text{4} & \text{1} & \text{2} & \text{3}  \\ \end{matrix} \right]$ and $\text{B=}\left[ \begin{matrix} \text{1} & \text{0} & \text{-2} & \text{1}  \\\end{matrix} \right]$ 

$\begin{align} & \text{AB=}\left[ \begin{matrix} \text{4} & \text{1} & \text{2} & \text{3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{0} & \text{-2} & \text{1}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{2} & \text{1} & \text{-4} & \text{3}  \\ \end{matrix} \right] \\ \end{align}$

(iv) ${{\mathbf{A}}^{2}}-\mathbf{AB}+2\mathbf{B}$ 

Ans:

$\text{A=}\left[ \begin{matrix} \text{4} & \text{1} & \text{2} & \text{3}  \\ \end{matrix} \right]$ and $\text{B=}\left[ \begin{matrix} \text{1} & \text{0} & \text{-2} & \text{1}  \\ \end{matrix} \right]$

${{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{18} & \text{7} & \text{14} & \text{11}  \\ \end{matrix} \right]$, $\text{AB=}\left[ \begin{matrix} \text{2} & \text{1} & \text{-4} & \text{3}  \\ \end{matrix} \right]$ and $\text{2B=}\left[ \begin{matrix} \text{2} & \text{0} & \text{-4} & \text{2}  \\\end{matrix} \right]$ 

${{\text{A}}^{\text{2}}}\text{-AB+2B=}\left[ \begin{matrix} \text{18} & \text{7} & \text{14} & \text{11}  \\ \end{matrix} \right]-\left[ \begin{matrix} \text{2} & \text{1} & \text{-4} & \text{3}  \\ \end{matrix} \right]+\left[ \begin{matrix} \text{2} & \text{0} & \text{-4} & \text{2}  \\\end{matrix} \right]$ $\text{=}\left[ \begin{matrix} \text{18} & \text{6} & \text{14} & \text{10}  \\ \end{matrix} \right]$ 

11. If $\text{A=}\left[ \begin{matrix} \mathbf{\text{1}} & \mathbf{\text{4}} & \mathbf{\text{1}} & \mathbf{\text{-3}}  \\ \end{matrix} \right]$ and $\text{B=}\left[ \begin{matrix} \mathbf{\text{1}} & \mathbf{\text{2}} &\mathbf{ \text{-1}} &\mathbf{ \text{-1}}  \\ \end{matrix} \right]$, find:

(i) ${{\left( A+B \right)}^{2}}$ 

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{1} & \text{4} & \text{1} & \text{-3}  \\ \end{matrix} \right]$ and $\text{B=}\left[ \begin{matrix} \text{1} & \text{2} & \text{-1} & \text{-1}  \\ \end{matrix} \right]$

Now, $\text{A+B=}\left[ \begin{matrix} \text{1} & \text{4} & \text{1} & \text{-3}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{1} & \text{2} & \text{-1} & \text{-1}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2} & \text{6} & \text{0} & \text{-4}  \\ \end{matrix} \right]$ 

${{\text{(A+B)}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{2} & \text{6} & \text{0} & \text{-4}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{6} & \text{0} & \text{-4}  \\ \end{matrix} \right]$ 

$\text{=}\left[ \begin{matrix} \text{4} & \text{-12} & \text{0} & \text{16}  \\\end{matrix} \right]$ 

(ii) A2+B2  

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{1} & \text{4} & \text{1} & \text{-3}  \\ \end{matrix} \right]$ and $\text{B=}\left[ \begin{matrix} \text{1} & \text{2} & \text{-1} & \text{-1}  \\ \end{matrix} \right]$

$\begin{align} & \text{Now,}\,{{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{1} & \text{4} & \text{1} & \text{-3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{4} & \text{1} & \text{-3}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{1} & \text{2} & \text{-1} & \text{-1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{2} & \text{-1} & \text{-1}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{5} & \text{-8} & \text{-2} & \text{13}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{-1} & \text{0} & \text{0} & \text{-1}  \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \text{4} & \text{-8} & \text{-2} & \text{12}  \\ \end{matrix} \right] \\ \end{align}$  

(iii) Is ${{\text{(A+B)}}^{\text{2}}}\text{=}{{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{?}$

Ans:

From (i) and (ii) it is clear that ${{\text{(A+B)}}^{\text{2}}}\ne {{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}$. 

12. Find the matrix A, if $B=\left[ 2~1~0~1~ \right]$ and ${{\text{B}}^{\text{2}}}\text{=B+}\dfrac{\text{1}}{\text{2}}\text{A}$. 

Ans:

Given, $\text{B=}\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{1}  \\ \end{matrix} \right]$ and ${{\text{B}}^{\text{2}}}\text{=B+}\dfrac{\text{1}}{\text{2}}\text{A}$ 

A[2(B2-B) 

$\begin{align} & \text{=2}\left( \left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{1}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{1}  \\ \end{matrix} \right] \right) \\  & \text{=2}\left( \left[ \begin{matrix} \text{4} & \text{3} & \text{0} & \text{1}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{1}  \\ \end{matrix} \right] \right) \\  & \text{=2}\left[ \begin{matrix} \text{2} & \text{2} & \text{0} & \text{0}  \\ \end{matrix} \right] \\  & \therefore \text{A=}\left[ \begin{matrix} \text{4} & \text{4} & \text{0} & \text{0}  \\ \end{matrix} \right] \\  \end{align}$ 

13. If $\text{A=}\left[ \begin{matrix} \mathbf{\text{-1}} & \mathbf{\text{1} }& \mathbf{\text{a}} & \mathbf{\text{b} } \\ \end{matrix} \right]$ and ${{\text{A}}^{\text{2}}}\text{=I}$, find a and b. 

Ans:

Given $\text{A=}\left[ \begin{matrix} \text{-1} & \text{1} & \text{a} & \text{b}  \\ \end{matrix} \right]$ and ${{\text{A}}^{\text{2}}}\text{=I}$ 

$\begin{align} & {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{-1} & \text{1} & \text{a} & \text{b}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{-1} & \text{1} & \text{a} & \text{b}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{1+a} & \text{b-1} & \text{-a+ab} & \text{a+}{{\text{b}}^{\text{2}}}  \\ \end{matrix} \right] \\ & \left[ \begin{matrix} \text{1+a} & \text{b-1} & \text{-a+ab} & a+{{b}^{\text{2}}}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right] \\ \end{align}$

By comparing we have

$\begin{align} & \text{1+a=1}\,\therefore \,\text{a=0} \\ & \text{b-1=0}\,\therefore \,\text{b=1} \\ \end{align}$ 

Thus the value of a=0 and b=1. 

14. If $\text{A=}\left[ \begin{matrix} \mathbf{\text{2}} & \mathbf{\text{1}} & \mathbf{\text{0}} & \mathbf{\text{0}}  \\ \end{matrix} \right]\text{,B=}\left[ \begin{matrix} \mathbf{\text{2}} & \mathbf{\text{3}} & \mathbf{\text{4}} &\mathbf{ \text{1}}  \\ \end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix} \mathbf{\text{1} }& \mathbf{\text{4}} & \mathbf{\text{0}} & \mathbf{\text{2}}  \\\end{matrix} \right]$; then show that:

(i) A(B+C)=AB+AC  

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{0}  \\ \end{matrix} \right]\text{,B=}\left[ \begin{matrix} \text{2} & \text{3} & \text{4} & \text{1}  \\ \end{matrix} \right]\,\,\text{and}\,\text{C=}\left[ \begin{matrix} \text{1} & \text{4} & \text{0} & \text{2}  \\ \end{matrix} \right]$

LHS = A(B+C) 

$\begin{align} & \text{=}\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{0}  \\ \end{matrix} \right]\left( \left[ \begin{matrix} \text{2} & \text{3} & \text{4} & \text{1}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{1} & \text{4} & \text{0} & \text{2}  \\ \end{matrix} \right] \right) \\  & \text{=}\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{0}  \\ \end{matrix} \right]\left( \left[ \begin{matrix} \text{3} & \text{7} & \text{4} & \text{3}  \\ \end{matrix} \right] \right) \\  & \text{=}\left[ \begin{matrix} \text{10} & \text{17} & \text{0} & \text{0}  \\ \end{matrix} \right] \\ \end{align}$

RHS = AB + AC

$\begin{align} & =\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{0}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{3} & \text{4} & \text{1}  \\ \end{matrix} \right]+\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{0}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{4} & \text{0} & \text{2}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{8} & \text{7} & \text{0} & \text{0}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{2} & \text{10} & \text{0} & \text{0}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{10} & \text{17} & \text{0} & \text{0}  \\ \end{matrix} \right] \\ & =LHS \\ \end{align}$ 

Hence Proved.

(ii) (B-A)C=BC-AC  

Ans: 

Given, $\text{A=}\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{0}  \\ \end{matrix} \right]\text{,B=}\left[ \begin{matrix} \text{2} & \text{3} & \text{4} & \text{1}  \\ \end{matrix} \right]\,\,\text{and}\,\text{C=}\left[ \begin{matrix} \text{1} & \text{4} & \text{0} & \text{2}  \\ \end{matrix} \right]$

LHS = (B-A)C

$=\left( \left[ \begin{matrix} \text{2} & \text{3} & \text{4} & \text{1}  \\ \end{matrix} \right]-\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{0}  \\ \end{matrix} \right] \right)\left[ \begin{matrix} \text{1} & \text{4} & \text{0} & \text{2}  \\ \end{matrix} \right]$ 

$\text{=}\left[ \begin{matrix} \text{0} & \text{2} & \text{4} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{4} & \text{0} & \text{2}  \\ \end{matrix} \right]$

$\text{=}\left[ \begin{matrix} \text{0} & \text{4} & \text{4} & \text{18}  \\ \end{matrix} \right]$

RHS = BC - AC 

$\begin{align} & =\left[ \begin{matrix}\text{2} & \text{3} & \text{4} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{4} & \text{0} & \text{2}  \\ \end{matrix} \right]-\left[ \begin{matrix} \text{2} & \text{1} & \text{0} & \text{0}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{4} & \text{0} & \text{2}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{2} & \text{14} & \text{4} & \text{18}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{10} & \text{0} & \text{0}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{0} & \text{4} & \text{4} & \text{18}  \\ \end{matrix} \right] \\  & =LHS \\  \end{align}$ 

Hence Proved.

15. If $\text{A=}\left[ \begin{matrix} \mathbf{\text{1}} & \mathbf{\text{4} }& \mathbf{\text{2}} & \mathbf{\text{1}}  \\ \end{matrix} \right]\text{,}\,\text{B=}\left[ \begin{matrix} \mathbf{\text{-3}} & \mathbf{\text{2}} & \mathbf{\text{4}} & \mathbf{\text{0}}  \\ \end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix} \mathbf{\text{1}} & \mathbf{\text{0}} & \mathbf{\text{0}} & \mathbf{\text{2}}  \\ \end{matrix} \right]$, simplify: A2+BC 

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{1} & \text{4} & \text{2} & \text{1}  \\ \end{matrix} \right]\text{,}\,\text{B=}\left[ \begin{matrix} \text{-3} & \text{2} & \text{4} & \text{0}  \\ \end{matrix} \right]\,\,\text{and}\,\text{C=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]$ 

${{\text{A}}^{\text{2}}}\text{+BC=}\left[ \begin{matrix} \text{1} & \text{4} & \text{2} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{4} & \text{2} & \text{1}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{-3} & \text{2} & \text{4} & \text{0}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]$ 

$\begin{align} & \text{=}\left[ \begin{matrix} \text{9} & \text{8} & \text{4} & \text{9}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{-3} & \text{4} & \text{4} & \text{0}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{6} & \text{12} & \text{8} & \text{9}  \\ \end{matrix} \right] \\ \end{align}$

16. Solve for x and y: 

(i) $\left[ \begin{matrix} \mathbf{\text{2} }& \mathbf{\text{5}} & \mathbf{\text{5}} & \mathbf{\text{2}}  \\ \end{matrix} \right]\left[ \begin{matrix} \mathbf{\text{x}} & \mathbf{\text{y}}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \mathbf{\text{-7}} &\mathbf{ \text{14}}  \\ \end{matrix} \right]$ 

Ans: 

Given 

$\begin{align} & \left[ \begin{matrix} \text{2} & \text{5} & \text{5} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{x} & \text{y}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{-7} & \text{14}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{2x+5y} & \text{5x+2y}  \\ \end{matrix} \right]\text{=}\,\left[ \begin{matrix} \text{-7} & \text{14}  \\ \end{matrix} \right] \\  \end{align}$ 

By comparing we have

$2\text{x}+5\text{y}=-7$    …(1)

$5\text{x}+2\text{y}=14$     …(2) 

By adding (1) and (2) we have

x+y=1             …(3)

By subtracting (1) from (2) we have

x-y=7              …(4)  

By solving (3) and (4) we have

x=4, y=-3 

(ii)  $\left[ \begin{matrix} \mathbf{\text{x+y}} & \mathbf{\text{x-4}}  \\ \end{matrix} \right]\left[ \begin{matrix} \mathbf{\text{-1}} & \mathbf{\text{-222} } \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \mathbf{\text{-7}} & \mathbf{\text{-11} } \\ \end{matrix} \right]$ 

Ans:

Given

$\begin{align} & \left[ \begin{matrix} \text{x+y} & \text{x-4}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{-1} & \text{-222}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{-7} & \text{-11}  \\ \end{matrix} \right] \\  & \left[ \text{-x-y+2x-8-2x-2y+2x-8} \right]\text{=}\left[ \begin{matrix} \text{-7} & \text{-11}  \\ \end{matrix} \right] \\  & \left[ \text{x-y-8-2y-8} \right]\text{=}\left[ \begin{matrix} \text{-7} & \text{-11}  \\ \end{matrix} \right] \\  \end{align}$

By comparing we have

$\begin{align} & \text{x-y-8=-7}\Rightarrow \text{x-y=1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\text{(1)} \\ & \text{-2y-8=-11}\Rightarrow \text{y=}\dfrac{\text{3}}{\text{2}} \\ \end{align}$

From (1), $\text{x=}\dfrac{\text{5}}{\text{2}}$ 

Thus value of $x=2.5$ and $y=1.5$ 

(iii) $\left[ \begin{matrix} \mathbf{\text{-2}} & \mathbf{\text{0} }& \mathbf{\text{3}} & \mathbf{\text{1}}  \\ \end{matrix} \right]\left[ \begin{matrix} \mathbf{\text{-12} }& \mathbf{\text{x}}  \\ \end{matrix} \right]\text{+3}\left[ \begin{matrix} \mathbf{\text{-2}} & \mathbf{\text{1}}  \\ \end{matrix} \right]\text{=2}\left[ \begin{matrix} \mathbf{\text{y}} & \mathbf{\text{3}}  \\ \end{matrix} \right]$ 

Ans:

Given 

$\begin{align} & \left[ \begin{matrix} -2 & 0 & 3 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   -12 & x  \\ \end{matrix} \right]+3\left[ \begin{matrix} -2 & 1  \\ \end{matrix} \right]=2\left[ \begin{matrix} y & 3  \\ \end{matrix} \right] \\   & \left[ \begin{matrix} -12 & x-3  \\ \end{matrix} \right]+\left[ \begin{matrix}  -6 & 3  \\ \end{matrix} \right]=\left[ \begin{matrix} 2y & 6  \\ \end{matrix} \right] \\  & \left[ \begin{matrix}  -4 & 2x  \\ \end{matrix} \right]=\left[ \begin{matrix} 2y & 6  \\ \end{matrix} \right] \\  \end{align}$ 

By comparing we have

$x=3$ and $y=-2$ 

17. In each case given below, find: 

1. The order of matrix M.

2. The matrix M.

(i) $\begin{align} & \text{M }\!\!\times\!\!\text{ }\left[ \begin{matrix}\mathbf{ \text{1} }& \mathbf{\text{1}} & \mathbf{\text{0} }& \mathbf{\text{2}}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \mathbf{\text{1}} & \mathbf{\text{2}}  \\ \end{matrix}\text{ }\!\!~\!\!\text{ } \right] \\  &  \\  \end{align}$

Ans:

Given 

$\text{M }\!\!\times\!\!\text{ }\left[ \begin{matrix} \text{1} & \text{1} & \text{0} & \text{2}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{1} & \text{2}  \\ \end{matrix}\text{ }\!\!~\!\!\text{ } \right]$ 

It is known that matrix multiplication is as

${{\text{A}}_{\text{l }\!\!\times\!\!\text{ n}}}\text{ }\!\!\times\!\!\text{ }{{\text{B}}_{\text{n }\!\!\times\!\!\text{ m}}}\text{=}{{\text{C}}_{\text{l }\!\!\times\!\!\text{ m}}}$ 

Here $A=M,~B=\left[ \begin{matrix} 1 & 1 & 0 & 2  \\ \end{matrix} \right]\,$ and $C=\left[ \begin{matrix} 1 & 2  \\ \end{matrix} \right]$

Order f B is $2\times 2$ order of C is $1\times 2$ 

It means order of M will be $1\times 2$ 

Let $M=\left[ \begin{matrix} a & b  \\ \end{matrix} \right]$ 

So, $\left[ \begin{matrix} a & b  \\ \end{matrix} \right]~\times \left[ \begin{matrix} \text{1} & \text{1} & \text{0} & \text{2}  \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 2  \\ \end{matrix} \right]$ 

$\left[ \begin{matrix} a & a+2b  \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 2  \\ \end{matrix} \right]$ 

By comparing we have 

$a=1$ and $b=\dfrac{1}{2}$ 

Matrix $M=\left[ \begin{matrix} 1 & \dfrac{1}{2}  \\\end{matrix} \right]$ 

(ii) $\left[ \begin{matrix} \mathbf{\text{1} }& \mathbf{\text{4}} & \mathbf{\text{2}} & \mathbf{\text{1}}  \\ \end{matrix} \right]\text{ }\!\!\times\!\!\text{ M=}\left[ \begin{matrix} \mathbf{\text{13}} & \mathbf{\text{5} } \\ \end{matrix}\text{ }\!\!~\!\!\text{ } \right]$ 

Ans:

Given, $\left[ \begin{matrix} 1 & 4 & 2 & 1  \\ \end{matrix} \right]\times M=\left[ \begin{matrix} 13 & 5  \\ \end{matrix}~ \right]$ 

It is known that matrix multiplication is as

${{A}_{l\times n}}\times {{B}_{n\times m}}={{C}_{l\times m}}$ 

Here $A=\left[ \begin{matrix} 1 & 4 & 2 & 1  \\ \end{matrix} \right],~B=M~and~C=\left[ \begin{matrix} 13 & 5  \\ \end{matrix}~ \right]$ 

Order f A is $2\times 2$ order of C is $2\times 1$ 

It means order of M will be $2\times 1$ 

Let $M=\left[ \begin{matrix} a & b  \\ \end{matrix} \right]$ 

So, $\left[ \begin{matrix} 1 & 4 & 2 & 1  \\ \end{matrix} \right]\times \left[ a~b~ \right]=\left[ \begin{matrix} 13 & 5  \\ \end{matrix}~ \right]$ 

$\left[ \begin{matrix} a+4b & 2a+b  \\ \end{matrix} \right]=\left[ \begin{matrix}  13 & 5  \\ \end{matrix}~ \right]$ 

By comparing we have 

$a+4b=13$   …(1) 

and $2a+b=5$      …(2) 

now, by $\left( 1 \right)\times 2-\left( 2 \right)$, we have

$7b=21\Rightarrow b=3$ 

From (1), $a=1$ 

Matrix $M=\left[ \begin{matrix} 1 & 3  \\ \end{matrix}~ \right]$ 

18. If $\text{A=}\left[ \begin{matrix} \mathbf{\text{2}} & \mathbf{\text{x}} & \mathbf{\text{0}} & \mathbf{\text{1}}  \\ \end{matrix} \right]$ and $\text{B=}\left[ \begin{matrix} \mathbf{\text{4} }& \mathbf{\text{3} }& \mathbf{\text{6}} & \mathbf{\text{0}} & \mathbf{\text{1} } \\ \end{matrix}\text{ }\!\!~\!\!\text{ } \right]$; find the values of x, given that ${{\text{A}}^{\text{2}}}\text{=B}$.

Ans:

Given, $A=\left[ \begin{matrix} 2 & x & 0 & 1  \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 4 & 3 & 6 & 0 & 1  \\ \end{matrix}~ \right]$ and

$\begin{align} & {{A}^{2}}=B\Rightarrow \left[ \begin{matrix} 2 & x & 0 & 1  \\ \end{matrix} \right]\left[ \begin{matrix} 2 & x & 0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix} 4 & 3 & 6 & 0 & 1  \\\end{matrix} \right] \\  & \left[ \begin{matrix} 4 & 3 & x & 0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix} 4 & 3 & 6 & 0 & 1  \\ \end{matrix} \right] \\  \end{align}$ 

By comparing we have

x=12

19. If $\text{A=}\left[ \begin{matrix} \mathbf{\text{3}} & \mathbf{\text{7}} & \mathbf{\text{2}} & \mathbf{\text{4}}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \mathbf{\text{0}} & \mathbf{\text{2}} & \mathbf{\text{5}} & \mathbf{\text{3}}  \\ \end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix} \mathbf{\text{1}} & \mathbf{\text{-5}} &\mathbf{ \text{-4}} & \mathbf{\text{6}}  \\ \end{matrix} \right]$. Find:$\text{AB-5C}$.

Ans:

Given that $A=\left[ \begin{matrix} 3 & 7 & 2 & 4  \\ \end{matrix} \right],~B=\left[ \begin{matrix} 0 & 2 & 5 & 3  \\ \end{matrix} \right]$$~A=\left[ 3~7~2~4~ \right],~B=\left[ 0~2~5~3~ \right]$ 

and $C=\left[ \begin{matrix} 1 & -5 & -4 & 6  \\ \end{matrix} \right]$

$\begin{align} & AB-5C=\left[ \begin{matrix} 3 & 7 & 2 & 4  \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 2 & 5 & 3  \\ \end{matrix} \right]-5\left[ \begin{matrix} 1 & -5 & -4 & 6  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} 35 & 27 & 20 & 16  \\ \end{matrix} \right]-\left[ \begin{matrix} 5 & -25 & -20 & 30  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} 30 & 52 & 40 & -14  \\ \end{matrix} \right] \\  \end{align}$

19. If A and B are any two $\text{2 }\!\!\times\!\!\text{ 2}$ matrices such that $\text{AB=BA=B}$ is not a zero matrix, what can you say about the matrix A?

Ans: 

Given that A and B are any two $2\times 2$ matrices such that $AB=BA=B\ne 0$. 

It is known that the identity matrix has a commutative property with any square matrix of the same order and gives the same matrix as a result.

$BI=IB=B$

Therefore, matrix A is an identity matrix of order $2\times 2$.

20. Given $\text{A=}\left[ \begin{matrix} \mathbf{\text{3}} & \mathbf{\text{0}} & \mathbf{\text{0}} &\mathbf{ \text{4}}  \\\end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \mathbf{\text{a}} & \mathbf{\text{b}} & \mathbf{\text{0} }& \mathbf{\text{c}}  \\ \end{matrix} \right]$ and that $\text{AB=A+B}$; find the values of $\text{a, }\!\!~\!\!\text{ b}$ and $\text{c}$. 

Ans:

Given $A=\left[ \begin{matrix} 3 & 0 & 0 & 4  \\ \end{matrix} \right],~B=\left[ \begin{matrix} a & b & 0 & c  \\ \end{matrix} \right]$ and

$AB=A+B$ 

$\begin{align} & \left[ \begin{matrix} 3 & 0 & 0 & 4  \\ \end{matrix} \right]\left[ \begin{matrix} a & b & 0 & c  \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & 0 & 0 & 4  \\ \end{matrix} \right]+\left[ \begin{matrix} a & b & 0 & c  \\ \end{matrix} \right] \\ & \left[ \begin{matrix} 3a & 3b & 0 & 4c  \\ \end{matrix} \right]=\left[ \begin{matrix} 3+a & b & 0 & 4+c  \\ \end{matrix} \right] \\ \end{align}$

By comparing we have

$3a=3+a\Rightarrow a=\dfrac{3}{2}$ 

$3b=b\Rightarrow b=0$ 

$4c=4+c\Rightarrow c=\dfrac{4}{3}$ 

Therefore we have values of $a=\dfrac{3}{2},~b=0~and~c=\dfrac{4}{3}$. 

21. If $\text{P=}\left[ \begin{matrix} \mathbf{\text{1}} &\mathbf{ \text{2} }& \mathbf{\text{2} }&\mathbf{ \text{-1}}  \\ \end{matrix} \right]$ and $\text{Q=}\left[ \begin{matrix} \mathbf{\text{1} }&\mathbf{ \text{0}} &\mathbf{ \text{2}} &\mathbf{ \text{1} } \\ \end{matrix} \right]$, then compute: 

(i) ${{\text{P}}^{\text{2}}}\text{-}{{\text{Q}}^{\text{2}}}$ 

Ans:

Given $P=\left[ \begin{matrix} 1 & 2 & 2 & -1  \\ \end{matrix} \right]$ and $Q=\left[ \begin{matrix} 1 & 0 & 2 & 1  \\ \end{matrix} \right]$  

$\begin{align} & {{P}^{2}}+{{Q}^{2}}=\left[ \begin{matrix} 1 & 2 & 2 & -1  \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 2 & -1  \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 0 & 2 & 1  \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2 & 1  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} 5 & 0 & 0 & 5  \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 0 & 4 & -1  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} 6 & 0 & 4 & 4  \\ \end{matrix} \right] \\ \end{align}$

(ii) $\left( \text{P+Q} \right)\left( \text{P-Q} \right)$ 

Ans:

Given $P=\left[ \begin{matrix} 1 & 2 & 2 & -1  \\ \end{matrix} \right]$ and $Q=\left[ \begin{matrix} 1 & 0 & 2 & 1  \\ \end{matrix} \right]$

$\begin{align} & (P+Q)(P-Q)=\left( \left[ \begin{matrix} 1 & 2 & 2 & -1  \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 2 & 2 & -1  \\ \end{matrix} \right] \right)\left( \left[ \begin{matrix} 1 & 2 & 2 & -1  \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 2 & 2 & -1  \\ \end{matrix} \right] \right) \\  & =\left[ \begin{matrix} 2 & 2 & 4 & 0  \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & 2 & 0 & -2  \\ \end{matrix} \right] \\   & =\left[ \begin{matrix} 0 & 0 & 0 & 8  \\ \end{matrix} \right] \\ \end{align}$ 

Therefore it is clear that $\left( \text{P+Q} \right)\left( \text{P-Q} \right)\ne {{\text{P}}^{\text{2}}}\text{-}{{\text{Q}}^{\text{2}}}$. 

So $\left( \text{P+Q} \right)\left( \text{P-Q} \right)\text{=}{{\text{P}}^{\text{2}}}\text{-}{{\text{Q}}^{\text{2}}}$ is not true for matrix.

22. Given the matrices: $\text{A=}\left[ \begin{matrix} \mathbf{\text{2}} &\mathbf{ \text{1}} & \mathbf{\text{4}} & \mathbf{\text{2}}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \mathbf{\text{3}} & \mathbf{\text{4}} & \mathbf{\text{-1}} & \mathbf{\text{-2}}  \\ \end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix} \mathbf{\text{-3}} & \mathbf{\text{1}} &\mathbf{ \text{0}} &\mathbf{ \text{-2}  }\\ \end{matrix} \right]$. Find:

(i) ABC

Ans:

Given $A=\left[ \begin{matrix} 2 & 1 & 4 & 2  \\ \end{matrix} \right],~B=\left[ \begin{matrix} 3 & 4 & -1 & -2  \\ \end{matrix} \right]$ and $C=\left[ \begin{matrix} -3 & 1 & 0 & -2  \\ \end{matrix} \right]$ 

$\begin{align} & ABC=\left[ \begin{matrix} 2 & 1 & 4 & 2  \\ \end{matrix} \right] \left[ \begin{matrix} 3 & 4 & -1 & -2  \\ \end{matrix} \right] \left[ \begin{matrix} -3 & 1 & 0 & -2  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} 5 & 6 & 10 & 12  \\ \end{matrix} \right] \left[ \begin{matrix} -3 & 1 & 0 & -2  \\ \end{matrix} \right] \\   & =\left[ \begin{matrix} -15 & -7 & -30 & -14  \\ \end{matrix} \right] \\ \end{align}$ 

(ii) ACB

Ans:

Given $A=\left[ \begin{matrix} 2 & 1 & 4 & 2  \\ \end{matrix} \right],~B=\left[ \begin{matrix} 3 & 4 & -1 & -2  \\ \end{matrix} \right]$ and $C=\left[ \begin{matrix} -3 & 1 & 0 & -2  \\ \end{matrix} \right]$ 

$ACB=\left[ \begin{matrix}   2 & 1 & 4 & 2\text{ }\!\!~\!\!\text{ }  \\\end{matrix} \right]\left[ \begin{matrix}   -3 & 1 & 0 & -2\text{ }\!\!~\!\!\text{ }  \\\end{matrix} \right]\left[ \begin{matrix}   3 & 4 & -1 & -2\text{ }\!\!~\!\!\text{ }  \\\end{matrix} \right]$

$\text{ }\!\!~\!\!\text{ }=\left[ \begin{matrix}   \text{ }\!\!~\!\!\text{ }-6 & 0 & -1 & 20\text{ }\!\!~\!\!\text{ }  \\\end{matrix} \right]\left[ \begin{matrix}   -3 & 1 & 0 & -2\text{ }\!\!~\!\!\text{ }  \\\end{matrix} \right]$

$\text{ }\!\!~\!\!\text{ }=\left[ \begin{matrix}   18 & -6 & 36 & -12\text{ }\!\!~\!\!\text{ }  \\\end{matrix} \right]$

Hence, $ABC\ne ACB$ 

23. If $\text{A=}\left[ \begin{matrix} \mathbf{\text{1}} &\mathbf{ \text{2}} &\mathbf{ \text{3}} & \mathbf{\text{4}}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \mathbf{\text{6}} & \mathbf{\text{1}} & \mathbf{\text{1}} & \mathbf{\text{1}}  \\ \end{matrix}\text{ }\!\!~\!\!\text{ } \right]$ and $\text{C=}\left[ \begin{matrix} \mathbf{\text{-2}} & \mathbf{\text{-3}} & \mathbf{\text{0}} &\mathbf{ \text{1}}  \\ \end{matrix} \right]$, find each of the following and state if they are equal:

(i) CA+B

Ans:

Given $A=\left[ \begin{matrix} 1 & 2 & 3 & 4  \\ \end{matrix} \right],~B=\left[ \begin{matrix} 6 & 1 & 1 & 1  \\ \end{matrix}~ \right]\,\,and\,C=\left[ \begin{matrix} -2 & -3 & 0 & 1  \\ \end{matrix}~ \right]$ 

$\begin{align} & CA+B=\left[ \begin{matrix} -2 & -3 & 0 & 1  \\ \end{matrix}~ \right]\left[ \begin{matrix} 1 & 2 & 3 & 4  \\ \end{matrix} \right]+\left[ \begin{matrix} 6 & 1 & 1 & 1  \\ \end{matrix}~ \right] \\  & =\left[ \begin{matrix} -11 & -16 & 3 & 4  \\ \end{matrix}~ \right]+\left[ \begin{matrix} 6 & 1 & 1 & 1  \\ \end{matrix}~ \right] \\  & =\left[ \begin{matrix} -5 & -15 & 4 & 5  \\ \end{matrix}~ \right] \\ \end{align}$

24. A+CB 

Ans:

Given $A=\left[ \begin{matrix} 1 & 2 & 3 & 4  \\ \end{matrix} \right],~B=\left[ \begin{matrix} 6 & 1 & 1 & 1  \\ \end{matrix}~ \right]\,\,and\,C=\left[ \begin{matrix} -2 & -3 & 0 & 1  \\ \end{matrix}~ \right]$  

$\begin{align} & A+CB=\left[ \begin{matrix} 1 & 2 & 3 & 4  \\ \end{matrix} \right]+\left[ \begin{matrix} -2 & -3 & 0 & 1  \\ \end{matrix}~ \right]\left[ \begin{matrix} 6 & 1 & 1 & 1  \\ \end{matrix}~ \right] \\  & =\left[ \begin{matrix} 1 & 2 & 3 & 4  \\ \end{matrix}~ \right]+\left[ \begin{matrix} -15 & -5 & 1 & 1  \\ \end{matrix}~ \right] \\  & =\left[ \begin{matrix} -14 & -3 & 4 & 5  \\ \end{matrix}~ \right] \\  \end{align}$

Hence they aren’t equal. 

25. If $\text{A=}\left[ \begin{matrix} \mathbf{\text{2}} &\mathbf{ \text{1}} & \mathbf{\text{1}} &\mathbf{ \text{3}}  \\ \end{matrix} \right]$ and $\text{B=}\left[ \begin{matrix} \mathbf{\text{3}} &\mathbf{ \text{-11}}  \\ \end{matrix} \right]$ find the value of matrix X such that $\text{AX=B}$

Ans:

Given $A=\left[ \begin{matrix} 2 & 1 & 1 & 3  \\ \end{matrix} \right]\,\,and\,B=\left[ \begin{matrix} 3 & -11  \\ \end{matrix} \right]$

We know that if matrix multiplication $AX=B$ is possible then order of X will be equal to order of B as matrix A is a square matrix, so matrix X can be taken as

$X=\left[ \begin{matrix} x & y  \\ \end{matrix} \right]$ 

So, $AX=B\Rightarrow \left[ \begin{matrix} 2 & 1 & 1 & 3  \\ \end{matrix} \right]\left[ \begin{matrix} x & y  \\ \end{matrix} \right]=\left[ 3~-11~ \right]$

$\left[ \begin{matrix} 2x+y & x+3y  \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & -11  \\ \end{matrix} \right]$  

Now by comparing we get

$2x+y=3$        …(1) 

$x+3y=-11$   …(2) 

By solving (1) and (2) we get

$x=4,~y=-5$ 

Hence matrix $X=\left[ 4~-5~ \right]$ 

26. If $\text{A=}\left[ \begin{matrix} \mathbf{\text{4}} &\mathbf{ \text{2}} &\mathbf{ \text{1}} &\mathbf{ \text{1}}  \\ \end{matrix}\text{ }\!\!~\!\!\text{ } \right]$. Find $\left( \text{A-2I} \right)\left( \text{A-3I} \right)$. 

Ans:

Given $A=\left[ \begin{matrix} 4 & 2 & 1 & 1  \\ \end{matrix} \right]$ and I is unit matrix so, $I=\left[ \begin{matrix} 1 & 0 & 0 & 1  \\ \end{matrix} \right]$ 

Now $\left( A-2I \right)\left( A-3I \right)=\left( \left[ \begin{matrix} 4 & 2 & 1 & 1  \\ \end{matrix} \right]-2\left[ \begin{matrix} 1 & 0 & 0 & 1  \\ \end{matrix}~ \right] \right)\left( \left[ \begin{matrix} 4 & 2 & 1 & 1  \\ \end{matrix} \right]-3\left[ \begin{matrix} 1 & 0 & 0 & 1  \\ \end{matrix}~ \right] \right)$ 

$=\left[ \begin{matrix} 2 & 2 & 1 & -1  \\ \end{matrix} \right] \left[ \begin{matrix} 1 & 2 & 1 & -2  \\ \end{matrix} \right]$ 

$=\left[ \begin{matrix} 4 & 0 & 0 & 4  \\ \end{matrix}~ \right]=4I$ 

$\left( A-2I \right)\left( A-3I \right)=4I$ 

27. If $\text{A=}\left[ \begin{matrix} \text{2} & \text{1} & \text{-1} & \text{0} & \text{1} & \text{-2}  \\ \end{matrix} \right]$, At is the transpose of matrix A, find:

(i) ${{\text{A}}^{\text{t}}}\text{A}$ 

Ans:

Given $A=\left[ \begin{matrix} 2 & 1 & -1 & 0 & 1 & -2  \\ \end{matrix} \right]$ then 

${{\text{A}}^{\text{t}}}\text{=}\left[ \begin{matrix} \text{2} & \text{0} & \text{1} & 1 & \text{-1} & \text{-2}  \\ \end{matrix} \right]$ 

${{\text{A}}^{\text{t}}}\text{A=}\left[ \begin{matrix} \text{2} & 0 & 1 & \text{1} & \text{-1} & \text{-2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{1} & \text{-1} & \text{0} & \text{1} & \text{-2}  \\ \end{matrix} \right]$

$=\left[ \begin{matrix} 4 & 2 & -2 & 2 & 2 & -3 & -2 & -3 & 5  \\ \end{matrix} \right]$

(ii) $\text{A}{{\text{A}}^{\text{t}}}$ 

Ans:

Given $A=\left[ \begin{matrix} 2 & 1 & -1 & 0 & 1 & -2  \\ \end{matrix} \right]$ then 

${{\text{A}}^{\text{t}}}\text{=}\left[ \begin{matrix} \text{2} & \text{0} & \text{1} & 1 & \text{-1} & \text{-2}  \\ \end{matrix} \right]$

$\text{A}{{\text{A}}^{\text{t}}}\text{=}\left[ \begin{matrix} \text{2} & \text{1} & \text{-1} & \text{0} & \text{1} & \text{-2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & 0 & 1 & \text{1} & \text{-1} & \text{-2}  \\ \end{matrix} \right]$

$=\left[ \begin{matrix} 6 & 3 & 3 & 5  \\ \end{matrix} \right]$ 

28. If $\text{M=}\left[ \begin{matrix} \mathbf{\text{4}} & \mathbf{\text{1}} & \mathbf{\text{-1}} & \mathbf{\text{2}}  \\ \end{matrix} \right]$, show that : $\text{6M-}{{\text{M}}^{\text{2}}}\text{=9I}$; where I is a $\text{2 }\!\!\times\!\!\text{ 2}$ unit matrix.

Ans:

Given $\text{M=}\left[ \begin{matrix} \text{4} & \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]$ 

Now, $\text{LHS}\,\text{=}\,\text{6M-}{{\text{M}}^{\text{2}}}$ 

$\text{=6}\left[ \begin{matrix} 4 & 1 & -1 & 2  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} 4 & 1 & -1 & 2  \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 1 & -1 & 2  \\ \end{matrix} \right]$ 

$\begin{align} & =\left[ \begin{matrix}  24 & 6 & -6 & 12  \\ \end{matrix} \right]-\left[ \begin{matrix} 15 & 6 & -6 & 3  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} 9 & 0 & 0 & 9  \\ \end{matrix} \right] \\  \end{align}$ 

$\begin{align} & \text{=9I} \\   & \text{=RHS} \\ \end{align}$ 

Hence Proved. 

29. If $\text{P=}\left[ \begin{matrix} \mathbf{\text{2}} & \mathbf{\text{6}} & \mathbf{\text{3}} & \mathbf{\text{9}}  \\ \end{matrix} \right]$ and $\text{Q=}\left[ \begin{matrix} \mathbf{\text{3}} & \mathbf{\text{x}} & \mathbf{\text{y}} & \mathbf{\text{2}}  \\ \end{matrix} \right]$, find x and y such that PQ = null matrix

Ans: 

Given $\text{P=}\left[ \begin{matrix} \text{2} & \text{6} & \text{3} & \text{9}  \\ \end{matrix} \right]$ and $\text{Q=}\left[ \begin{matrix} \text{3} & \text{x} & \text{y} & \text{2}  \\ \end{matrix} \right]$

$\begin{align} & \text{PQ}=0\Rightarrow \left[ 2~6~3~9~ \right]\left[ 3~x~y~2~ \right]=0 \\  & \left[ \begin{matrix} 6+6y & 2x+12 & 9+9y & 3x+18  \\ \end{matrix} \right]=0 \\  & \left[ \begin{matrix} 6(1+y) & 2(x+6) & 9(1+y) & 3(x+6)  \\ \end{matrix} \right]=0 \\ \end{align}$ 

By comparing we have 

x = -6 and y = -1.

30. Evaluate: 

$\mathbf{\left[ \text{2}\,\text{cos }\!\!~\!\!\text{ }\,\text{60 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ }\,\text{-}\,\text{2 }\!\!~\!\!\text{ sin}\,\text{30 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ }\,\text{-}\,\text{tan }\!\!~\!\!\text{ }\,\text{45 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ }\,\text{cos }\!\!~\!\!\text{ }\,\text{0 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ } \right]\left[ \text{cot }\!\!~\!\!\text{ }\,\text{45 }\!\!{}^\circ\!\!\text{ }\,\text{cosec }\!\!~\!\!\text{ }\,\text{30 }\!\!{}^\circ\!\!\text{ }\,\text{sec }\!\!~\!\!\text{ }\,\text{60 }\!\!{}^\circ\!\!\text{ }\,\text{sin }\!\!~\!\!\text{ 90 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ } \right]}$ 

Ans:

Given 

$\left[ \text{2}\,\text{cos }\!\!~\!\!\text{ }\,\text{60 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ }\,\text{-}\,\text{2 }\!\!~\!\!\text{ sin}\,\text{30 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ }\,\text{-}\,\text{tan }\!\!~\!\!\text{ }\,\text{45 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ }\,\text{cos }\!\!~\!\!\text{ }\,\text{0 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ } \right]\left[ \text{cot }\!\!~\!\!\text{ }\,\text{45 }\!\!{}^\circ\!\!\text{ }\,\text{cosec }\!\!~\!\!\text{ }\,\text{30 }\!\!{}^\circ\!\!\text{ }\,\text{sec }\!\!~\!\!\text{ }\,\text{60 }\!\!{}^\circ\!\!\text{ }\,\text{sin }\!\!~\!\!\text{ 90 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ } \right]$

$=\left[ \begin{matrix}   2\times \frac{1}{2} & -2\times \frac{1}{2} & -1 & 1\text{ }\!\!~\!\!\text{ }  \\\end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 2 & 1\text{ }\!\!~\!\!\text{ }  \\\end{matrix} \right]$

$\text{ }\!\!~\!\!\text{ }=\left[ \begin{matrix}   1 & -1 & -1 & 1\text{ }\!\!~\!\!\text{ }  \\\end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 2 & 1\text{ }\!\!~\!\!\text{ }  \\\end{matrix} \right]$

$\text{ }\!\!~\!\!\text{ }=\left[ \begin{matrix}   -1 & 1 & 1 & -1\text{ }\!\!~\!\!\text{ }  \\\end{matrix} \right]$

31. State, with reason, whether the following are true or false. A, B and C are matrices of order $\text{2 }\!\!\times\!\!\text{ 2}$. 

(i) A+B=B+A

Ans:

Given: A, B and C are matrices of order $\text{2 }\!\!\times\!\!\text{ 2}$ 

Matrices follow the commutative rule for addition. 

Therefore, A+B=B+A, is true.

(ii)A-B=B-A

Ans: 

Given: A, B and C are matrices of order $\text{2 }\!\!\times\!\!\text{ 2}$

Matrices do not follow the commutative rule for subtraction. 

Therefore, $A-B=B-A$ is false.

(iii) $\left( \text{BC} \right)\text{A=B}\left( \text{CA} \right)$

Ans: 

Given: A, B and C are matrices of order $\text{2 }\!\!\times\!\!\text{ 2}$

Matrices do not follow the commutative rule for multiplication. 

Therefore, $\left( BC \right)A=B\left( CA \right)$ is false.

(iv) $\left( \text{A+B} \right)\text{C=AC+BC}$ 

Ans: 

Given: A, B and C are matrices of order $\text{2 }\!\!\times\!\!\text{ 2}$

Matrices follow the distributive rule. 

Therefore, $\left( A+B \right)C=AC+BC$  is true.

(v) $\text{A}\left( \text{B-C} \right)\text{=AB-AC}$ 

Ans: 

Given: A, B and C are matrices of order $\text{2 }\!\!\times\!\!\text{ 2}$  

Matrices follow the distributive rule. 

Therefore, $A\left( B-C \right)=AB-AC$ is true.

(vi) $\left( \text{A-B} \right)\text{C=AC-BC}$  

Ans: 

Given: A, B and C are matrices of order $\text{2 }\!\!\times\!\!\text{ 2}$

Matrices follow the distributive rule. 

Therefore, $\left( A-B \right)C=AC-BC$  is true.

(vii) ${{\text{A}}^{\text{2}}}\text{-}{{\text{B}}^{\text{2}}}\text{=}\left( \text{A+B} \right)\left( \text{A-B} \right)$ 

Ans: 

Given: A, B and C are matrices of order $\text{2 }\!\!\times\!\!\text{ 2}$

Matrices follow the distributive rule but not commutative rule over multiplication. 

Therefore, ${{A}^{2}}-{{B}^{2}}=\left( A+B \right)\left( A-B \right)$  is false.

(viii) ${{\left( \text{A-B} \right)}^{\text{2}}}\text{=}{{\text{A}}^{\text{2}}}\text{-2AB+}{{\text{B}}^{\text{2}}}$ 

Ans: 

Given: A, B and C are matrices of order $\text{2 }\!\!\times\!\!\text{ 2}$

Matrices follow the distributive rule but not commutative rule over multiplication. 

${{\left( A-B \right)}^{2}}={{A}^{2}}-AB-BA+{{B}^{2}}$ 

Therefore,$~{{\left( A-B \right)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}$   is false.

Exercise-9 D

1. Find x and y, if: $\left[ \begin{matrix} \mathbf{\text{3}} &\mathbf{ \text{-2}} & \mathbf{\text{-1}} &\mathbf{ \text{4}}  \\ \end{matrix} \right]\left[ \begin{matrix} \mathbf{\text{2x}} &\mathbf{ \text{1}}  \\ \end{matrix} \right]\text{+2}\left[ \begin{matrix} \mathbf{\text{-4}} & \mathbf{\text{5}}  \\ \end{matrix} \right]\text{=4}\left[ \begin{matrix} \mathbf{\text{2}} & \mathbf{\text{y}}  \\ \end{matrix} \right]$ 

Ans:

Given $\left[ \begin{matrix} \text{3} & \text{-2} & \text{-1} & \text{4}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2x} & \text{1}  \\ \end{matrix} \right]\text{+2}\left[ \begin{matrix} \text{-4} & \text{5}  \\ \end{matrix} \right]\text{=4}\left[ \begin{matrix} \text{2} & \text{y}  \\ \end{matrix} \right]$ 

$\begin{align} & \left[ \begin{matrix} \text{6x-2} & \text{-2x+4}  \\ \end{matrix} \right]+\left[ \begin{matrix} \text{-8} & \text{10}  \\ \end{matrix} \right]=\left[ \begin{matrix} \text{8} & \text{4y}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{6x-10} & \text{-2x+14}  \\ \end{matrix} \right]=\left[ \begin{matrix} \text{8} & \text{4y}  \\ \end{matrix} \right] \\  \end{align}$ 

By comparing we have 

$\text{6x-10=8}\Rightarrow \text{x=3}$ 

$\text{-2x+14=4y}\Rightarrow \text{y=2}$ 

Therefore the value of $x=3$ and $y=2$. 

2. Find x and y if: $\left[ \begin{matrix} \mathbf{\text{3x}} & \mathbf{\text{8}}  \\  \end{matrix} \right]\left[ \begin{matrix} \mathbf{\text{1}} & \mathbf{\text{4}} & \mathbf{\text{3}} & \mathbf{\text{7}}  \\ \end{matrix} \right]\text{-3}\left[ \begin{matrix} \mathbf{\text{2}} & \mathbf{\text{-7}}  \\ \end{matrix} \right]\,\text{=}\,\text{5}\left[ \begin{matrix} \mathbf{\text{3}} & \mathbf{\text{2y}}  \\ \end{matrix} \right]$ 

Ans:

Given,

$\left[ \begin{matrix} \text{3x} & \text{8}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{4} & \text{3} & \text{7}  \\ \end{matrix} \right]\text{-3}\left[ \begin{matrix} \text{2} & \text{-7}  \\ \end{matrix} \right]\,\text{=}\,\text{5}\left[ \begin{matrix} \text{3} & \text{2y}  \\ \end{matrix} \right]$ 

$\begin{align} & \left[ \begin{matrix} \text{3x+24} & \text{12x+56}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{6} & \text{-21}  \\ \end{matrix} \right]\,\text{=}\,\left[ \begin{matrix} \text{15} & \text{10}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{3x+18} & \text{12x+77}  \\ \end{matrix} \right]\,\text{=}\,\left[ \begin{matrix} \text{15} & \text{10y}  \\ \end{matrix} \right] \\  \end{align}$ 

By comparing we have

$\text{3x+18=15}\Rightarrow \text{x=-1}$ 

$\text{12x+77=10y}\Rightarrow \text{y=}\dfrac{\text{65}}{\text{10}}\text{=}\dfrac{\text{13}}{\text{2}}\text{=6}\text{.5}$ 

Therefore the value of x = -1 and y = 6.5. 

3. If $\left[ \begin{matrix} \mathbf{\text{x}} & \mathbf{\text{y} } \\ \end{matrix} \right]\left[ \begin{matrix} \mathbf{\text{x} }& \mathbf{\text{y}}  \\ \end{matrix} \right]\text{=}\left[ 25 \right]$ and $\left[ \begin{matrix} \mathbf{\text{-x}} & \mathbf{\text{y} } \\ \end{matrix} \right]\left[ \begin{matrix} \mathbf{\text{2x}} &\mathbf{ \text{y} } \\ \end{matrix} \right]\text{=}\left[ \text{-2} \right]$; find x and y, if :

(i) $x,~y\in W$ (whole number) 

Ans: 

Given,

$\left[ \begin{matrix} \text{x} & \text{y}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{x} & \text{y}  \\ \end{matrix} \right]\text{=}\left[ \text{25} \right]$ 

$\left[ {{x}^{2}}+{{y}^{2}} \right]=\left[ \text{25} \right]$ 

$\therefore {{x}^{2}}+{{y}^{2}}\text{=25}$   …(1)

and 

$\left[ \begin{matrix}  -x & y  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2}x & y  \\ \end{matrix} \right]\text{=}\left[ \text{-2} \right]$ 

$\left[ \text{-2}{{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right]\text{=}\left[ \text{-2} \right]$ 

$\therefore \,\text{2}{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=2}$   …(2) 

By adding (1) and (2) we get that

$\text{3}{{\text{x}}^{\text{2}}}\text{=27}\Rightarrow {{\text{x}}^{\text{2}}}\text{=9}$ 

$\therefore x=\pm \text{3}$ but $x\in W\therefore x=\text{3}$ 

${{\text{y}}^{\text{2}}}\text{=25-9=16}$ 

$y=\pm \text{4}$ but $y\in W\therefore y=\text{4}$ 

Thus the value of $x=\text{3}$ and y = 4 

(ii) $x,~y\in Z$ (integers)

Ans:

Given,

$\left[ \begin{matrix} \text{x} & \text{y}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{x} & \text{y}  \\ \end{matrix} \right]\text{=}\left[ \text{25} \right]$ 

$\left[ {{x}^{2}}+{{y}^{2}} \right]=\left[ \text{25} \right]$ 

$\therefore {{x}^{2}}+{{y}^{2}}\text{=25}$   …(1)

And

$\left[ \begin{matrix} -x & y  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2}x & y  \\ \end{matrix} \right]\text{=}\left[ \text{-2} \right]$ 

$\left[ \text{-2}{{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right]\text{=}\left[ \text{-2} \right]$ 

$\therefore \,\text{2}{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=2}$   …(2)  

By adding (1) and (2) we get that

$\text{3}{{x}^{2}}=\text{27}\Rightarrow {{x}^{2}}=\text{9}$ 

$\therefore x=\pm \text{3}$ 

${{y}^{2}}\text{=25-9=16}$ 

$y=\pm \text{4}$ 

Thus, the value of $x=\pm \text{3}$ and $y=\pm \text{4}$ 

4. Given, $\left[ \begin{matrix} \mathbf{2} & \mathbf{1 }& \mathbf{-3} & \mathbf{4}  \\ \end{matrix} \right]X=\left[ \begin{matrix} \mathbf{7} & \mathbf{6}  \\ \end{matrix} \right]$. Write:

(i) The order of the matrix X.

Ans:

Given, $\left[ \begin{matrix} \text{2} & \text{1} & \text{-3} & \text{4}  \\ \end{matrix} \right]\text{X=}\left[ \begin{matrix} \text{7} & \text{6}  \\ \end{matrix} \right]$   

It is known that matrix multiplication is as

${{A}_{l\times n}}\times {{B}_{n\times m}}={{C}_{l\times m}}$ 

Here $A\,=\,\left[ \begin{matrix} \text{2} & \text{1} & \text{-3} & \text{4}  \\ \end{matrix} \right]\text{,}\,\text{B=X}$ and $\text{C}\,\text{=}\,\left[ \begin{matrix} \text{7} & \text{6}  \\ \end{matrix} \right]$  

Order of A is $\text{2 }\!\!\times\!\!\text{ 2}$ order of C is $\text{2 }\!\!\times\!\!\text{ 1}$ 

It means the order of X will be $\text{2 }\!\!\times\!\!\text{ 1}$. 

(ii) The matrix X. 

Ans:

Given, $\left[ \begin{matrix} \text{2} & \text{1} & \text{-3} & \text{4}  \\ \end{matrix} \right]\text{X=}\left[ \begin{matrix} \text{7} & \text{6}  \\ \end{matrix} \right]$   

Matrix X can be taken as $X=\left[ \begin{matrix} x & y  \\ \end{matrix} \right]$ 

$\left[ \begin{matrix} \text{2} & \text{1} & \text{-3} & \text{4}  \\ \end{matrix} \right]\left[ \begin{matrix} x & y  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{7} & \text{6}  \\ \end{matrix} \right]$

$\left[ \begin{matrix} \text{2x+y} & \text{-3x+4y}  \\ \end{matrix}\text{ }\!\!~\!\!\text{ } \right]\text{=}\left[ \begin{matrix} \text{7} & \text{6}  \\ \end{matrix} \right]$ 

By comparing we have

2x+y=7        …(1) 

$\text{-3x+4y=6}$   …(2) 

By $\left( \text{1} \right)\,\text{ }\!\!\times\!\!\text{ }\,\text{4-}\left( \text{2} \right)$ we have

$\text{11}x=\text{22}\therefore x=\text{2}$ 

$\text{y=7-8=-1}$ 

Thus, matrix $\text{X=}\left[ \text{2 }\!\!~\!\!\text{ -1 }\!\!~\!\!\text{ } \right]$ 

5. Evaluate:

$\left[ \begin{matrix} \mathbf{cos~45{}^\circ}  & \mathbf{sin30{}^\circ}  & \mathbf{\sqrt{2}~cos~0{}^\circ}  & \mathbf{sin~0{}^\circ ~}  \\ \end{matrix} \right]\left[ \begin{matrix} \mathbf{sin~45{}^\circ}  & \mathbf{cos~90{}^\circ}  & \mathbf{sin~90{}^\circ}  & \mathbf{cot~45{}^\circ}   \\ \end{matrix} \right]$ 

Ans:

Given 

$\left[ \begin{matrix} cos~45{}^\circ  & sin30{}^\circ  & \sqrt{2}~cos~0{}^\circ  & sin~0{}^\circ ~  \\ \end{matrix} \right]\left[ \begin{matrix} sin~45{}^\circ  & cos~90{}^\circ  & sin~90{}^\circ  & cot~45{}^\circ   \\ \end{matrix} \right]$

$\begin{align} & \text{=}\left[ \begin{matrix} \dfrac{\text{1}}{\sqrt{\text{2}}} & \dfrac{\text{1}}{\text{2}} & \sqrt{\text{2}} & \text{0}  \\ \end{matrix} \right]\left[ \begin{matrix} \dfrac{\text{1}}{\sqrt{\text{2}}} & \text{0} & \text{1} & \text{1}  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} \dfrac{\text{1}}{\text{2}}+\dfrac{\text{1}}{\text{2}} & \text{0+}\dfrac{\text{1}}{\text{2}} & \text{1} & \text{0}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{1} & \dfrac{\text{1}}{\text{2}} & \text{1} & \text{0}  \\ \end{matrix} \right] \\  \end{align}$ 

6. If $A=\left[ \begin{matrix} \mathbf{0} & \mathbf{-1} & \mathbf{4 }& \mathbf{-3}  \\ \end{matrix} \right],~B=\left[ \begin{matrix} \mathbf{-5} & \mathbf{6}  \\ \end{matrix} \right]$ and $3A\times M=2B$; find matrix M. 

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{0} & \text{-1} & \text{4} & \text{-3}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \text{-5} & \text{6}  \\ \end{matrix} \right]$  

$\text{3}A\times M=\text{2}B$ 

Matrix M can be taken as, $M=\left[ x~y~ \right]$ 

$\begin{align} & \text{3}\left[ \begin{matrix} \text{0} & \text{-1} & \text{4} & \text{-3}  \\ \end{matrix} \right]\left[ \begin{matrix} x & y  \\ \end{matrix} \right]\text{=2}\left[ \begin{matrix} \text{-5} & \text{6}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{-3y} & \text{12x-9y}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{-10} & \text{12}  \\ \end{matrix} \right] \\ \end{align}$

By comparing

$\text{-3y=-10}\therefore \text{y=}\dfrac{\text{10}}{\text{3}}$ 

$\text{12x-9y=12}$ 

$\text{x=}\dfrac{\text{42}}{\text{12}}\text{=}\dfrac{\text{7}}{\text{2}}$ 

Matrix, $M=\left[ \begin{matrix} \dfrac{\text{7}}{\text{2}} & \dfrac{\text{10}}{\text{3}}  \\ \end{matrix}\text{ }\!\!~\!\!\text{ } \right]$ 

7. If $\left[ \begin{matrix} \mathbf{a} &\mathbf{ 3 }&\mathbf{ 4} &\mathbf{ 1}  \\ \end{matrix} \right]+\left[ \begin{matrix} \mathbf{2} &\mathbf{ b} &\mathbf{ 1} &\mathbf{ -2}  \\ \end{matrix} \right]-\left[ \begin{matrix} \mathbf{1} & \mathbf{1} &\mathbf{ -2} &\mathbf{ c}  \\ \end{matrix} \right]=\left[ \begin{matrix} \mathbf{5} &\mathbf{ 0} &\mathbf{ 7} &\mathbf{ 3}  \\ \end{matrix} \right]$, find the value of a, b and c. 

Ans:

Given, $\left[ \begin{matrix} \text{a} & \text{3} & \text{4} & \text{1}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{2} & \text{b} & \text{1} & \text{-2}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{1} & \text{1} & \text{-2} & \text{c}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{5} & \text{0} & \text{7} & \text{3}  \\ \end{matrix} \right]$  

$\left[ \begin{matrix} \text{a+2-1} & \text{3+b-1} & \text{4+1+2} & \text{1-2-c}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{5} & \text{0} & \text{7} & \text{3}  \\ \end{matrix} \right]$ 

By comparing we have

$a+\text{1}=\text{5}\therefore a=\text{4}$ 

$b+\text{2=0}\therefore b=\text{-2}$ 

$-\text{1}-c=\text{3}\therefore c=-\text{4}$ 

Thus the value of  $a=\text{4},~b=-\text{2}$ and c = -4.

8. If $A=\left[ \begin{matrix} \mathbf{1} & \mathbf{2 }& \mathbf{2} & \mathbf{1}  \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} \mathbf{2} &\mathbf{ 1} &\mathbf{ 1} &\mathbf{ 2}  \\ \end{matrix} \right]$ ; find : 

(i) $\mathbf{A\left( BA \right)}$

Ans: 

Given, $A=\left[ \begin{matrix} 1 & 2 & 2 & 1  \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 2 & 1 & 1 & 2  \\ \end{matrix} \right]$

$A\left( BA \right)=\left[ \begin{matrix} 1 & 2 & 2 & 1  \\ \end{matrix} \right]\left( \left[ \begin{matrix} 1 & 2 & 2 & 1  \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 1 & 1 & 2  \\ \end{matrix} \right] \right)$ 

$=\left[ \begin{matrix} 1 & 2 & 2 & 1  \\ \end{matrix} \right]\left[ \begin{matrix} 4 & \text{5} & 5 & 4  \\ \end{matrix} \right]$ 

$=\left[ \begin{matrix} 14 & 13 & 13 & 1  \\\end{matrix}\text{4} \right]$ 

(ii) $\mathbf{\left( AB \right)B}$ 

Ans: 

Given, $A=\left[ \begin{matrix} 1 & 2 & 2 & 1  \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 2 & 1 & 1 & 2  \\ \end{matrix} \right]$ 

$\left( \text{AB} \right)\text{B=}\left( \left[ \begin{matrix} \text{1} & \text{2} & \text{2} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{1} & \text{1} & \text{2}  \\ \end{matrix} \right] \right)\left[ \begin{matrix} \text{2} & \text{1} & \text{1} & \text{2}  \\ \end{matrix} \right]$ 

$\begin{align} & \text{=}\left[ \begin{matrix} \text{4} & \text{5} & \text{5} & \text{4}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{1} & \text{1} & \text{2}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{13} & \text{14} & \text{14} & \text{13}  \\ \end{matrix} \right] \\  \end{align}$

9. Find x and y, if: $\left[ \begin{matrix} \mathbf{x} & \mathbf{3x} & \mathbf{y} & \mathbf{4y}  \\ \end{matrix} \right]\left[ \begin{matrix} \mathbf{2} & \mathbf{1}  \\ \end{matrix} \right]=\left[ \begin{matrix} \mathbf{5} & \mathbf{12}  \\ \end{matrix} \right]$. 

Ans:

Given,

$\left[ \begin{matrix} x & 3x & y & 4y  \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 12  \\ \end{matrix} \right]$ 

$\left[ \begin{matrix} \text{5x} & \text{6y}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{5} & \text{12}  \\ \end{matrix} \right]$ 

By comparing we have,

$x=\text{1},~y=\text{2}$.

10. If matrix $X=\left[ \begin{matrix} \mathbf{-3} & \mathbf{4 }& \mathbf{2} & \mathbf{-3}  \\ \end{matrix} \right]\left[ \begin{matrix} \mathbf{2} & \mathbf{-2}  \\ \end{matrix} \right]$ and $2X-3Y=\left[ \begin{matrix} \mathbf{10} & \mathbf{-8}  \\ \end{matrix} \right]$, find the matrix $'X'$ and matrix $'{Y}'$. 

Ans: 

Given 

$X=\left[ \begin{matrix} -3 & 4 & 2 & -3  \\ \end{matrix} \right]\left[ \begin{matrix} 2 & -2  \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 4 & 1 & 0  \\ \end{matrix} \right]$

and $\text{2X-3Y=}\left[ \begin{matrix}  \text{10} & \text{-8}  \\ \end{matrix} \right]$ 

$\text{3Y=2X-}\left[ \begin{matrix} \text{10} & \text{-8}  \\ \end{matrix} \right]\text{=2}\left[ \begin{matrix} \text{-14} & \text{10}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{10} & \text{-8}  \\ \end{matrix} \right]$

$\text{Y=}\dfrac{\text{1}}{\text{3}}\left[ \begin{matrix} \text{-38} & \text{28}  \\ \end{matrix} \right]$ 

$\text{Y=}\left[ \begin{matrix} \text{-}\dfrac{\text{38}}{\text{3}} & \dfrac{\text{28}}{\text{3}}  \\ \end{matrix} \right]$ 

11. Given $A=\left[ \begin{matrix} \mathbf{2} & \mathbf{-1} &\mathbf{ 2} &\mathbf{ 0}  \\ \end{matrix} \right],~B=\left[ \begin{matrix} \mathbf{-3} &\mathbf{ 2} &\mathbf{ 4} & \mathbf{0 } \\ \end{matrix} \right]$ and $C=\left[ \begin{matrix} \mathbf{1} &\mathbf{ 0} &\mathbf{ 0} &\mathbf{ 2}  \\ \end{matrix} \right]$, find the matrix X such that

(i) $A+X=2B+C$ .

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{2} & \text{-1} & \text{2} & \text{0}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \text{-3} & \text{2} & \text{4} & \text{0}  \\ \end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]$ 

$A+X=\text{2}B+C$ 

$X=\text{2}B+C-A$ 

$\text{=2}\left[ \begin{matrix} \text{-3} & \text{2} & \text{4} & \text{0}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{2} & \text{-1} & \text{2} & \text{0}  \\ \end{matrix} \right]$ 

$\text{=}\left[ \begin{matrix} \text{-6+1-2} & \text{4+0+1} & \text{8+0-2} & \text{0+0-0 }\!\!~\!\!\text{ }  \\\end{matrix} \right]$ 

$\text{X=}\left[ \begin{matrix} \text{-7} & \text{5} & \text{6} & \text{0}  \\\end{matrix} \right]$ 

12. Find the value of x, given that ${{A}^{2}}=B$, $A=\left[ \begin{matrix} \mathbf{2} & \mathbf{12 }& \mathbf{0} &\mathbf{ 1}  \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} \mathbf{4} &\mathbf{ x} &\mathbf{ 0} &\mathbf{ 1}  \\\end{matrix} \right]$ 

Ans:

Given $\text{A=}\left[ \begin{matrix} \text{2} & \text{12} & \text{0} & \text{1}  \\ \end{matrix} \right]$ and $\text{B=}\left[ \begin{matrix} \text{4} & \text{x} & \text{0} & \text{1}  \\ \end{matrix} \right]$ 

${{A}^{2}}=B$ 

$\begin{align} & \left[ \begin{matrix} \text{2} & \text{12} & \text{0} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{12} & \text{0} & \text{1}  \\ \end{matrix} \right]=\left[ \begin{matrix} \text{4} & \text{x} & \text{0} & \text{1}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{4} & \text{36} & \text{0} & \text{1}  \\ \end{matrix} \right]=\left[ \begin{matrix} \text{4} & \text{x} & \text{0} & \text{1}  \\ \end{matrix} \right] \\ \end{align}$ 

Thus, the value of x = 36. 

13. If $A=\left[ \begin{matrix} \mathbf{2} &\mathbf{ 5 }& \mathbf{1} &\mathbf{ 3}  \\ \end{matrix} \right],~B=\left[ \begin{matrix} \mathbf{4} & \mathbf{-2} &\mathbf{ -1} &\mathbf{ 3}  \\ \end{matrix} \right]$ and I is the identity matrix of the same order and At is the transpose of matrix A, find ${{A}^{t}}B+BI$. 

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{2} & \text{5} & \text{1} & \text{3}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \text{4} & \text{-2} & \text{-1} & \text{3}  \\ \end{matrix} \right]$

Transpose matrix of A, ${{\text{A}}^{\text{t}}}\text{=}\left[ \begin{matrix} \text{2} & \text{5} & \text{1} & \text{3}  \\ \end{matrix} \right]$ 

${{A}^{t}}B+BI=\left[ \begin{matrix}  2 & 5 & 1 & 3  \\\end{matrix} \right]\left[ \begin{matrix}   4 & -2 & -1 & 3  \\\end{matrix} \right]+\left[ \begin{matrix}   4 & -2 & -1 & 3  \\\end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 0 & 1  \\\end{matrix} \right]$

$=\left[ \begin{matrix}   7 & -1 & 17 & -1  \\\end{matrix} \right]+\left[ \begin{matrix}   4 & -2 & -1 & 3  \\\end{matrix} \right]$

$=\left[ \begin{matrix}   11 & -3 & -1 & 2  \\\end{matrix} \right]$

14. Given $A=\left[ \begin{matrix} \mathbf{2} & \mathbf{-6} & \mathbf{2} & \mathbf{0}  \\ \end{matrix} \right],~B=\left[ \begin{matrix} \mathbf{-3} & \mathbf{2} & \mathbf{4 }& \mathbf{0}  \\ \end{matrix} \right]$ and $C=\left[ \begin{matrix} \mathbf{4} & \mathbf{0 }& \mathbf{0} &\mathbf{ 2 } \\ \end{matrix} \right]$. Find the matrix X such that $A+2X=2B+C$ .

Ans:

Given $\text{A=}\left[ \begin{matrix} \text{2} & \text{-6} & \text{2} & \text{0}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \text{-3} & \text{2} & \text{4} & \text{0}  \\ \end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix} \text{4} & \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]$

$\text{A+2X=2B+C}$ 

$\text{2X=2B+C-A}$ 

$\text{X=}\frac{\text{1}}{\text{2}}\left( \text{2}\left[ \begin{matrix}   \text{-3} & \text{2} & \text{4} & \text{0}  \\\end{matrix} \right]\text{+}\left[ \begin{matrix}   \text{4} & \text{0} & \text{0} & \text{2}  \\\end{matrix} \right]\text{-}\left[ \begin{matrix}   \text{2} & \text{-6} & \text{2} & \text{0}  \\\end{matrix} \right] \right)$

$\text{=}\frac{\text{1}}{\text{2}}\left[ \begin{matrix}   \text{-6+4-2} & \text{4+0+6} & \text{8+0-2} & \text{0+2-0}  \\\end{matrix} \right]$

$\text{X=}\left[ \begin{matrix}   \text{-}\frac{\text{4}}{\text{2}} & \frac{\text{10}}{\text{2}} & \frac{\text{6}}{\text{2}} & \frac{\text{2}}{\text{2}}  \\\end{matrix} \right]\text{=}\left[ \begin{matrix}   \text{-2} & \text{5} & \text{3} & \text{1}  \\\end{matrix} \right]$

15. Let $A=\left[ \begin{matrix} \mathbf{4} & \mathbf{-2} &\mathbf{ 6 }&\mathbf{ -3}  \\ \end{matrix} \right],~B=\left[ \begin{matrix} \mathbf{0} & \mathbf{2} & \mathbf{1} &\mathbf{ -1}  \\ \end{matrix} \right]$ and $C=\left[ \begin{matrix} -2 & 3 & 1 & -1  \\ \end{matrix} \right]$. Find ${{A}^{2}}-A+BC$. 

Ans:

Given, $\text{A=}\left[ \begin{matrix} \text{4} & \text{-2} & \text{6} & \text{-3}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \text{0} & \text{2} & \text{1} & \text{-1}  \\ \end{matrix} \right]$ and $\text{C=}\left[ \begin{matrix} \text{-2} & \text{3} & \text{1} & \text{-1}  \\ \end{matrix} \right]$

$\begin{align} & {{\text{A}}^{\text{2}}}\text{-A+BC=}\left[ \begin{matrix} \text{4} & \text{-2} & \text{6} & \text{-3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{4} & \text{-2} & \text{6} & \text{-3}  \\ \end{matrix} \right]-\left[ \begin{matrix} \text{4} & \text{-2} & \text{6} & \text{-3}  \\ \end{matrix} \right]+\left[ \begin{matrix} \text{0} & \text{2} & \text{1} & \text{-1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{-2} & \text{3} & \text{1} & \text{-1}  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} \text{4} & \text{-2} & \text{6} & \text{-3}  \\ \end{matrix} \right]-\left[ \begin{matrix} \text{4} & \text{-2} & \text{6} & \text{-3}  \\ \end{matrix} \right]+\left[ \begin{matrix} \text{2} & \text{-2} & \text{-3} & \text{4}  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} \text{2} & \text{-2} & \text{-3} & \text{4}  \\ \end{matrix} \right] \\ \end{align}$

16. Let $A=\left[ \begin{matrix} \mathbf{1} & \mathbf{0} & \mathbf{2} & \mathbf{1}  \\ \end{matrix} \right],\,B=\left[ \begin{matrix} \mathbf{2} &\mathbf{ 3} &\mathbf{ -1} & \mathbf{0}  \\ \end{matrix} \right]$. Find ${{A}^{2}}+AB+{{B}^{2}}$ 

Ans: 

Given, $\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{2} & \text{1}  \\\end{matrix} \right]\text{,}\,\text{B=}\left[ \begin{matrix} \text{2} & \text{3} & \text{-1} & \text{0}  \\ \end{matrix} \right]$

$\begin{align} & {{\text{A}}^{\text{2}}}\text{+AB+}{{\text{B}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{1} & \text{0} & \text{2} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{0} & \text{2} & \text{1}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{1} & \text{0} & \text{2} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{3} & \text{-1} & \text{0}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{2} & \text{3} & \text{-1} & \text{0}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{3} & \text{-1} & \text{0}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{1} & \text{0} & \text{4} & \text{1}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{2} & \text{3} & \text{3} & \text{6}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{1} & \text{6} & \text{-2} & \text{-3}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{1+2+1} & \text{0+3+6} & \text{4+3-2} & \text{1+6-3}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{4} & \text{9} & \text{5} & \text{4}  \\ \end{matrix} \right] \\  \end{align}$ 

17. If $A=\left[ \begin{matrix} \mathbf{3} & \mathbf{a} & \mathbf{-4} & \mathbf{8}  \\ \end{matrix} \right],~B=\left[ \begin{matrix} \mathbf{c} & \mathbf{4} & \mathbf{-3 }& \mathbf{0}  \\ \end{matrix} \right],~C=\left[ \begin{matrix} \mathbf{-1} & \mathbf{4} & \mathbf{3} & \mathbf{b}  \\ \end{matrix} \right]$ and $3A-2C=6B$, find the values of a, b and c. 

Ans:

Given $\text{A=}\left[ \begin{matrix} \text{3} & \text{a} & \text{-4} & \text{8}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \text{c} & \text{4} & \text{-3} & \text{0}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ C=}\left[ \begin{matrix} \text{-1} & \text{4} & \text{3} & \text{b}  \\ \end{matrix} \right]$ 

$\text{3A-2C=6B}$ 

$\begin{align} & \text{3}\left[ \begin{matrix} \text{3} & \text{a} & \text{-4} & \text{8}  \\ \end{matrix} \right]\text{-2}\left[ \begin{matrix} \text{-1} & \text{4} & \text{3} & \text{b}  \\ \end{matrix} \right]\text{=6}\left[ \begin{matrix} \text{c} & \text{4} & \text{-3} & \text{0}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{9} & \text{3a} & \text{-12} & \text{24}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{-2} & \text{8} & \text{6} & \text{b}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{6c} & \text{24} & \text{-18} & \text{0}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{11} & \text{3a-8} & \text{-18} & \text{24-2b}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{6c} & \text{24} & \text{-18} & \text{0}  \\ \end{matrix} \right] \\  \end{align}$

By comparing we have

$\text{11=6c}\Rightarrow \text{c=}\dfrac{\text{11}}{\text{6}}$ 

$\text{3a-8=24}\Rightarrow \text{a=}\dfrac{\text{32}}{\text{3}}$ 

$\text{24-2b=0}\Rightarrow \text{b=12}$ 

Hence value of $\text{a=}\dfrac{\text{32}}{\text{3}}\text{, }\!\!~\!\!\text{ b=12}$ and $\text{c=}\dfrac{\text{11}}{\text{6}}$. 

18. Given $A=\left[ \begin{matrix} \mathbf{p} & \mathbf{0} & \mathbf{0} & \mathbf{2}  \\ \end{matrix} \right],~B=\left[ \begin{matrix} \mathbf{0} & \mathbf{-q} &\mathbf{ 1} &\mathbf{ 0}  \\ \end{matrix} \right],~C=\left[ \begin{matrix} \mathbf{2} & \mathbf{-2} & \mathbf{2} & \mathbf{2}  \\ \end{matrix} \right]$ and $BA={{C}^{2}}$. Find the value of p and q. 

Ans:

Given $\text{A=}\left[ \begin{matrix} \text{p} & \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \text{0} & \text{-q} & \text{1} & \text{0}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ C=}\left[ \begin{matrix} \text{2} & \text{-2} & \text{2} & \text{2}  \\ \end{matrix} \right]$ and $BA={{C}^{2}}$ 

$\begin{align} & \left[ \begin{matrix} \text{0} & \text{-q} & \text{1} & \text{0}  \\\end{matrix} \right]\left[ \begin{matrix} \text{p} & \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2} & \text{-2} & \text{2} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{-2} & \text{2} & \text{2}  \\ \end{matrix} \right] \\   & \left[ \begin{matrix} \text{0} & \text{-2q} & \text{p} & \text{0}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{0} & \text{-8} & \text{8} & \text{0}  \\ \end{matrix} \right] \\ \end{align}$

By comparing we have

p = 8 and q = 4

19. Given $A=\left[ \begin{matrix} \mathbf{3} & \mathbf{-2} & \mathbf{-1} &\mathbf{ 4}  \\ \end{matrix} \right],~B=\left[ \begin{matrix} \mathbf{6} & \mathbf{1}  \\ \end{matrix} \right],~C=\left[ \begin{matrix} \mathbf{-4} & \mathbf{5}  \\ \end{matrix} \right]$ and $D=\left[ \begin{matrix} \mathbf{2} & \mathbf{2}  \\ \end{matrix} \right]$. 

Find: AB+2C-4D. 

Ans: 

Given $\text{A=}\left[ \begin{matrix} \text{3} & \text{-2} & \text{-1} & \text{4}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \text{6} & \text{1}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ C=}\left[ \begin{matrix} \text{-4} & \text{5}  \\ \end{matrix} \right]$ and $\text{D=}\left[ \begin{matrix} \text{2} & \text{2}  \\ \end{matrix} \right]$

$\begin{align} & \text{AB+2C-4D=}\left[ \begin{matrix} \text{3} & \text{-2} & \text{-1} & \text{4}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{6} & \text{1}  \\ \end{matrix} \right]+\text{2}\left[ \begin{matrix} \text{-4} & \text{5}  \\ \end{matrix} \right]-\text{4}\left[ \begin{matrix} \text{2} & \text{2}  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} \text{16} & \text{-2}  \\ \end{matrix} \right]+\left[ \begin{matrix} \text{-8} & \text{10}  \\ \end{matrix} \right]-\left[ \begin{matrix} \text{8} & \text{8}  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} \text{16-8-8} & \text{-2+10-8}  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} \text{0} & \text{0}  \\ \end{matrix} \right] \\  \end{align}$

20. Evaluate: $\left[ \begin{matrix} \mathbf{4sin~30{}^\circ}  & \mathbf{2~cos~60{}^\circ}  & \mathbf{sin~90{}^\circ}  & \mathbf{2~cos~0{}^\circ ~}  \\ \end{matrix} \right]\left[ \begin{matrix} \mathbf{4} & \mathbf{5} & \mathbf{5 }& \mathbf{4}  \\ \end{matrix} \right]$. 

Ans:

Given, $\left[ \begin{matrix} \text{4sin }\!\!~\!\!\text{ 30 }\!\!{}^\circ\!\!\text{ } & \text{2 }\!\!~\!\!\text{ cos }\!\!~\!\!\text{ 60 }\!\!{}^\circ\!\!\text{ } & \text{sin }\!\!~\!\!\text{ 90 }\!\!{}^\circ\!\!\text{ } & \text{2 }\!\!~\!\!\text{ cos }\!\!~\!\!\text{ 0 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ }  \\ \end{matrix} \right]\left[ \begin{matrix} \text{4} & \text{5} & \text{5} & \text{4}  \\ \end{matrix} \right]$  

$\begin{align} & \text{=}\left[ \begin{matrix} \text{4 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}} & \text{2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}} & \text{1} & \text{2 }\!\!\times\!\!\text{ 1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{4} & \text{5} & \text{5} & \text{4}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{2} & \text{1} & \text{1} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{4} & \text{5} & \text{5} & \text{4}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{13} & \text{14} & \text{14} & \text{13}  \\ \end{matrix} \right] \\  \end{align}$ 

21. If $A=\left[ \begin{matrix} \mathbf{3} & \mathbf{1} & \mathbf{-1} &\mathbf{ 2}  \\ \end{matrix} \right]$ and I is unit matrix, find ${{A}^{2}}-5A+7I$. 

Ans: 

Given, ${{A}^{2}}-\text{5}A+7I$ 

$\begin{align} & \text{=}\left[ \begin{matrix} \text{3} & \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3} & \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\text{-5}\left[ \begin{matrix} \text{3} & \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\text{+7}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{8} & \text{5} & \text{-5} & \text{3}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{15} & \text{5} & \text{-5} & \text{10}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{7} & \text{0} & \text{0} & \text{7}  \\ \end{matrix} \right] \\  \end{align}$

$\begin{align} & \text{=}\left[ \begin{matrix} \text{8-15+7} & \text{5-5+0} & \text{-5+5+0} & \text{3-10+7}  \\ \end{matrix} \right] \\  & \text{=}\left[ \begin{matrix} \text{0} & \text{0} & \text{0} & \text{0}  \\ \end{matrix} \right] \\  \end{align}$

22. Given $A=\left[ \begin{matrix} \mathbf{2} & \mathbf{0} & \mathbf{-1} &\mathbf{ 7}  \\ \end{matrix} \right]$ and $I=\left[ \begin{matrix} \mathbf{1} &\mathbf{ 0} &\mathbf{ 0 }& \mathbf {1}  \\ \end{matrix} \right]$ and ${{A}^{2}}=9A+mI$. Find m. 

Ans:

Given $\text{A=}\left[ \begin{matrix} \text{2} & \text{0} & \text{-1} & \text{7}  \\ \end{matrix} \right]$ and $\text{I=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]$ 

${{A}^{2}}=9A+mI$ 

$\begin{align} & \left[ \begin{matrix} \text{2} & \text{0} & \text{-1} & \text{7}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{0} & \text{-1} & \text{7}  \\ \end{matrix} \right]=\text{9}\left[ \begin{matrix} \text{2} & \text{0} & \text{-1} & \text{7}  \\ \end{matrix} \right]+m\left[ \begin{matrix} \text{1} & \text{0} & \text{0} & \text{1}  \\\end{matrix} \right] \\  & \left[ \begin{matrix} \text{4} & \text{0} & \text{-9} & \text{49}  \\ \end{matrix} \right]=\left[ \begin{matrix} \text{18} & \text{0} & \text{-9} & \text{63}  \\ \end{matrix} \right]+\left[ \begin{matrix} \text{m} & \text{0} & \text{0} & m  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{4} & \text{0} & \text{-9} & \text{49}  \\ \end{matrix} \right]=\left[ \begin{matrix} \text{18+m} & \text{0} & \text{-9} & \text{63+m}  \\ \end{matrix} \right] \\  \end{align}$

By comparing we get

$\text{4=18+m}\Rightarrow \text{m=-14}$ 

$\text{49=63+m}\Rightarrow \text{m=-14}$ 

From both the caparisons we have m = -14. 

23. Given matrix $A=\left[ 4~sin~30{}^\circ ~cos~0{}^\circ ~cos~0{}^\circ ~4~sin~30{}^\circ ~ \right]$ and $B=\left[ \begin{matrix}  \mathbf{4} & \mathbf{5}  \\ \end{matrix} \right]$ . If $AX=B$ 

(i) Write the order of matrix X. 

Ans:

Given $\text{A=}\left[ \text{4 }\!\!~\!\!\text{ sin }\!\!~\!\!\text{ 30 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ cos }\!\!~\!\!\text{ 0 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ cos }\!\!~\!\!\text{ 0 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ 4 }\!\!~\!\!\text{ sin }\!\!~\!\!\text{ 30 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ } \right]$ and $\text{B=}\left[ \begin{matrix} \text{4} & \text{5}  \\ \end{matrix} \right]$ 

$AX=B$ 

The order of matrix X is $\text{2 }\!\!\times\!\!\text{ 2}$, as this is the only way to define the multiplication in the given equation.

(ii) Find the matrix X. 

Ans: 

Given  $\text{A=}\left[ \text{4 }\!\!~\!\!\text{ sin }\!\!~\!\!\text{ 30 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ cos }\!\!~\!\!\text{ 0 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ cos }\!\!~\!\!\text{ 0 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ 4 }\!\!~\!\!\text{ sin }\!\!~\!\!\text{ 30 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ } \right]$ and $\text{B=}\left[ \begin{matrix} \text{4} & \text{5}  \\ \end{matrix} \right]$

$AX=B$ 

Let matrix $X=\left[ \begin{matrix} p & q  \\ \end{matrix} \right]$ 

$\left[ \begin{matrix} \text{4 }\!\!~\!\!\text{ sin }\!\!~\!\!\text{ 30 }\!\!{}^\circ\!\!\text{ } & \text{cos }\!\!~\!\!\text{ 0 }\!\!{}^\circ\!\!\text{ } & \text{cos }\!\!~\!\!\text{ 0 }\!\!{}^\circ\!\!\text{ } & \text{4 }\!\!~\!\!\text{ sin }\!\!~\!\!\text{ 30 }\!\!{}^\circ\!\!\text{ }  \\ \end{matrix} \right]\left[ \begin{matrix} p & q  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{4} & \text{5}  \\ \end{matrix} \right]$ 

$\begin{align} & \left[ \begin{matrix} \text{4 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}} & \text{1} & \text{1} & \text{4 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{p} & \text{q}  \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ =}\left[ \begin{matrix} \text{4} & \text{5}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{2} & \text{1} & \text{1} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{p} & \text{q}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{4} & \text{5}  \\ \end{matrix} \right] \\   & \left[ \begin{matrix} \text{2p+q} & \text{p+2q}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{4} & \text{5}  \\ \end{matrix} \right] \\ \end{align}$ 

By comparing we have

$\text{2p+q=4}$   …(1) 

$\text{p+2q=5}$   …(2) 

By $\left( \text{1} \right)\text{ }\!\!\times\!\!\text{ -}\left( \text{2} \right)$,  $\text{4p-p=8-5}\Rightarrow \text{p=1}$ 

From (1), $\text{q=4-2 }\!\!\times\!\!\text{ 1=2}$ 

Thus the matrix $\text{X=}\left[ \begin{matrix} \text{1} & \text{2}  \\ \end{matrix} \right]$ 

24. If $A=\left[ \begin{matrix} \mathbf{1} & \mathbf{3} & \mathbf{3} & \mathbf{4}  \\ \end{matrix} \right],~B=\left[ \begin{matrix} \mathbf{-2} & \mathbf{1} & \mathbf{-3} & \mathbf{2}  \\ \end{matrix} \right]$ and ${{A}^{2}}-5{{B}^{2}}=5C$. Find matrix C, where C is a $2\times 2$ matrix.

Ans:

Given $\text{A=}\left[ \begin{matrix} \text{1} & \text{3} & \text{3} & \text{4}  \\ \end{matrix} \right]\text{, }\!\!~\!\!\text{ B=}\left[ \begin{matrix} \text{-2} & \text{1} & \text{-3} & \text{2}  \\ \end{matrix} \right]$ and ${{A}^{2}}-\text{5}{{B}^{2}}=\text{5}C$ 

Let matrix $C=\left[ \begin{matrix} p & q & r & s  \\ \end{matrix} \right]$ 

${{A}^{2}}-\text{5}{{B}^{2}}=\text{5}C$ 

$\begin{align} & \left[ \begin{matrix} \text{1} & \text{3} & \text{3} & \text{4}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{3} & \text{3} & \text{4}  \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ -5}\left[ \begin{matrix}  \text{-2} & \text{1} & \text{-3} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{-2} & \text{1} & \text{-3} & \text{2}  \\ \end{matrix} \right]\text{=5}\left[ \begin{matrix} \text{p} & \text{q} & \text{r} & \text{s}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{10} & \text{15} & \text{15} & \text{25}  \\ \end{matrix} \right]\text{-5}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]\text{=5}\left[ \begin{matrix} \text{p} & \text{q} & \text{r} & \text{s}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{1} & \text{3} & \text{3} & \text{4}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{p} & \text{q} & \text{r} & \text{s}  \\ \end{matrix} \right] \\  \end{align}$

Thus matrix $C=\left[ \begin{matrix} \text{1} & \text{3} & \text{3} & \text{4}  \\\end{matrix} \right]$ 

25. Given matrix $B=\left[ \begin{matrix} \mathbf{1} & \mathbf{1} & \mathbf{8} & \mathbf{3}  \\ \end{matrix} \right]$. Find the matrix X, if $X={{B}^{2}}-4B$. Hence, solve for a and b given $X\left[ \begin{matrix} \mathbf{a} & \mathbf{b}  \\ \end{matrix} \right]=\left[ \begin{matrix} \mathbf{5} & \mathbf {50}  \\\end{matrix} \right]$. 

Ans: 

Given $\text{B=}\left[ \begin{matrix} \text{1} & \text{1} & \text{8} & \text{3}  \\ \end{matrix} \right]$ and $X={{B}^{2}}-\text{4}B$ 

$\begin{align} & X=\left[ \begin{matrix} \text{1} & \text{1} & \text{8} & \text{3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{1} & \text{8} & \text{3}  \\ \end{matrix} \right]-\text{4}\left[ \begin{matrix} \text{1} & \text{1} & \text{8} & \text{3}  \\ \end{matrix} \right] \\  & =\left[ \begin{matrix} \text{9} & \text{4} & \text{32} & \text{17}  \\ \end{matrix} \right]-\left[ \begin{matrix} \text{1} & \text{1} & \text{8} & \text{3}  \\ \end{matrix} \right] \\  & X=\left[ \begin{matrix} \text{8} & \text{3} & \text{24} & \text{14}  \\ \end{matrix} \right] \\ \end{align}$ 

Now, 

$\begin{align} & \left[ \begin{matrix} \text{8} & \text{3} & \text{24} & \text{14}  \\ \end{matrix} \right]\left[ \begin{matrix} a & b  \\ \end{matrix} \right]=\left[ \begin{matrix} \text{5} & \text{50}  \\ \end{matrix} \right] \\  & \left[ \begin{matrix} \text{8a+3b} & \text{24a+14b}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{5} & \text{50}  \\ \end{matrix} \right] \\ \end{align}$

By comparing we have

$\text{8a+3b=5}$       …(1) 

$\text{12a+7b=25}$    …(2) 

By $\left( \text{1} \right)\text{ }\!\!\times\!\!\text{ 7-}\left( \text{2} \right)\text{ }\!\!\times\!\!\text{ 3}$ we have

$\text{56a-36a=35-75}$ 

$\text{20a=-40}\Rightarrow \text{a=-2}$ 

From (1), $\text{b=}\dfrac{\text{1}}{\text{3}}\left( \text{5-8 }\!\!\times\!\!\text{ }\left( \text{-2} \right) \right)\text{=7}$ 

Hence, the value of $\text{a}\,\text{=}\,\text{-2}$ and . ....

Order of the Matrices 

The order of the Matrices  having m rows and n columns will be m*n.  Matrices are generally represented by capital letters and their order.

Rows of the Matrices 

The horizontal lines in the Matrices  are called the rows of the Matrices . 

Columns of Matrices 

The columns are an arrangement of numbers in a vertical line of matrices.

Type of Matrices

Row Matrices : The row Matrices  are the Matrices  having a single row. For example, $\begin{bmatrix}1 &2  &3 \end{bmatrix}$

Column Matrices : A column Matrix  is a Matrix  having a single column.

Square Matrix: Square matrices are the matrices that have the number of rows and number of columns equal.

Rectangular Matrices: The matrices in which the number of rows is not equal to the number of rows.

Zero Matrices:  Zero matrices are matrices that have each element as 0. 

Diagonal Matrices: The diagonal matrix  is the matrix  having all of the elements as 0 except the elements of diagonal.

Identity Matrices : The identity matrices are matrices having all other elements as 0 and diagonal elements as 1. 

Transpose of a Matrices: The transpose of a matrix  is obtained by interchanging its rows with its columns.

Equality of Matrices : Two matrices are equal only if :

1. Both of the matrices have the same order.

2. Corresponding elements of both of the matrices are the same.

Addition of Matrices : Two matrices can be only added if they have the same order. You can simply add the matrices by adding simply their corresponding terms.

Subtraction of Matrices : You can only subtract matrices if they have the same order. To subtract a Matrix  from another, you can just simply subtract it from the corresponding terms of another one.

If A, B, and C are matrices which have same order, then : 

1. Matrix  Addition is commutative.

           A+ B = B+ A 

2. Matrix  addition is associative.

A+ (B+C)= ( A+B)+C 

3. A+ X = B 

X= B- A

Additive Identity of Matrices : Additive identity is the term by which adding any number will be as it is.0 is the additive identity of every number. Similarly, the additive identity of the Matrices  is a thanks Matrices .

Additive Inverse of a Matrices : The additive inverse of a Matrices  is the Matrices  after which adding you get the identity Matrices . So,the identity Matrices  of A will be - A.

If A+ B = B+ A= 0, then A is the additive inverse of B and B is the additive inverse of A. 

Multiplication of Matrices  by a scalar : A scalar is here a real number. 

Case 1:

You can simply multiply a Matrices  by a scalar just multiplying each of the elements of the Matrices  by the scaler 

Case 2:

 You can simply multiply two matrices A and B

 only if the number of columns in A is equal to the number of rows in B. If A has an order m*n then B should have ordered a*b then the AB Matrices  is possible only If n = a. 

Also, the order of Matrices  AB will be m*b.

For example, if A Matrices  has order 2*3 anhashave order 3*4, then AB will have order 2*4.

Conclusion

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